Text

Attachment to Response to NRC Request for Additional Information Re Containment Structure Conformance to Design Basis Requirements. Solvia Engineering Report SE 99-5, Attachment 6 to AEP:NRC:2520, Pages 1 - B29.6
	ML022030415
Person / Time
Site:	Cook
Issue date:	07/16/2002
From:	Greenlee S; Indiana Michigan Power Co
To:	; Document Control Desk, Office of Nuclear Reactor Regulation
References
	AEP:NRC:2520, TAC MB3603, TAC MB3604 SE 99-5
	Download: ML022030415 (163)
	v • d • e

Stv n o soits10 State Street, H /WI a~wu s0] 80l Sni 617 932 9580 Fax 617 933 4428 A structural-mechanical consulting engineeringfirm BOSTON.CLFETLE-[ND July 2, 2002 American Electric Power Nuclear Generation Group 500 Circle Drive Buchanan, MI 49107

Subject:

License for SOLVIA Computer Code Let this letter serve as a license for American Electric Power Company to use SOLVIA copyrighted verification documentation1 for the purposes of regulatory compliance in answering of questions as posed by the USNRC. This grants American Electric Power the right have a copy of this documentation and to send a copy of the same to the USNRC for their use in the review of the SOLVIA code.

The SOLVIA verification documents transmitted herewith each bear a SOLVIA copyright notice. The NRC is permitted to make the number of copies of the information contained in these reports which are necessary for its internal use in connection with generic and plant specific reviews and approvals as well as the issuance, denial, amendment, transfer, renewal, modification, suspension, revocation or violation of a license, permit, order, or regulation, copyright protection notwithstanding. The NRC is permitted to make the number of copies beyond those necessary for its internal use which are necessary in order to have one copy available for public viewing in the appropriate docket files in the public document room in Washington, DC and in local public document rooms as may be required by NRC regulations if the number of copies submitted is insufficient for this purpose. Copies made by the NRC must include the copyright notice in all instances.

Stevenson & Associates hereby grants this license as US distributor of the SOLVIA code.

Very truly yo rs, W Iter Dj rdjevic a) SOLVIA Finite Element Program, Version 99.0, SOLVIA Verification Manual Linear Examples, Report SE 99-4, SOLVIA Engineering AB, 2000.

b) SOLVIA Finite Element Program, Version 99.0, SOLVIA Verification Manual Nonlinear Examples, Report SE 99-5, SOLVIA Engineering AB, 2000.

ATTACHMENT 6 TO AEP:NRC:2520 SOLVIA ENGINEERING REPORT SE 99-5, "SOLVIA VERIFICATION MANUAL NON-LINEAR EXAMPLES"

SOLVIA. Finite Element System Version 99.0 SOLVIA Verification Manual Nonlinear Examples Report SE 99-5 SOLVIA Engineering AB

SOLVIA Verification Manual, Nonlinear Examples.

This manual, as well as the software described in it, is furnished under license and may be used or copied only in accordance with the terms of such license. The content of this manual is furnished for informational use only, is subject to change without notice, and should not be construed as a commitment by SOLVIA Engineering AB. SOLVIA Engineering AB assumes no responsibility or liability for any errors, inaccuracies or defects that may appear in this manual.

Except as permitted by such license, no part of this manual may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, recording, or otherwise, without the prior written permission of SOLVIA Engineering AB.

SOLVIA is a trademark or registered trademark of SOLVIA Engineering AB. Other tradenames, trademarks and registered trademarks included herein are property of their respective holders.

SOLVIA Engineering AB Tel +46-21-144050 Trefasgatan 3 Fax +46-21-188890 SE-721 30 Vdsterfls engineering@solvia.se Sweden www.solvia.com

SOLVIA Verification Manual INTRODUCTORY REMARKS The objective with this report is to present example solutions obtained with the SOLVIA-PRE, SOLVIA and SOLVIA-POST computer programs (the SOLVIA System) that verify and demonstrate their usage. Solutions to nonlinear analyses are presented in this report. Linear example solutions are presented in the companion report SE 99-4.

Since the aim is to compare the analysis results with analytical solutions, relatively small problems are solved, that also allow insight into the results. The analyses reported upon can be directly rerun with version 99.0 of the SOLVIA System. Complete input data for SOLVIA-PRE and SOLVIA POST is given for each example. All plot pictures created by the input data have been output to Microsoft Word in the PostScript graphical language.

We intend to update this report with further example solutions as we continue our work on the SOLVIA System. If you have any suggestions regarding the example solutions presented in this manual or suggestions on additional problems, we would be glad to hear from you.

Version 99.0 Nonlinear Examples I

SOLVIA Verification Manual CONTENTS B I Elastic-plastic analysis of TRUSS structure B2 Large amplitude oscillation of a pendulum B3 Elastic buckling of a cylindrical panel B4 Elastic buckling of an imperfect square plate B5 Cantilever beam in creep B6 Pipe whip analysis, direct integration solution B7 Pipe whip analysis, mode superposition B8 Cyclic loading of plastic truss, isotropic hardening B9 Cyclic loading of plastic truss, kinematic hardening B 10 Cable under gravity load B 11 Rubber sheet in large strain analysis B 12 Large displacement analysis of a spherical shell B 13 Elastic-plastic analysis of a thick-walled cylinder B 14 Thermo-plastic analysis of a thick-walled cylinder B 15 Cyclic creep analysis of a thick-walled cylinder B 16 Large displacement analysis of a plate B 17 Thermo-elastic analysis of a cantilever beam B 18 Analysis of an underground opening B 19 Reinforced concrete beam, PLANE STRESS B20 Concrete material curves in compression B21 Large deflection analysis of a shallow arch B22 Eight-story building subjected to impact load B23 Analysis of a sand specimen under compression B24 Cantilever under large displacements, SHELL B25 Cantilever under large displacements, PLATE B26 Plastic cantilever under pure bending, SHELL B27 Plastic cantilever under pure bending, PLATE B28 Transient temperature analysis of a slab B29 Transient temperature analysis of a space shuttle B30 Solidification of a semi-infinite slab of liquid B31 Solidification of a comer region B32 Excavation of a tunnel B33 Cable/frame structure with cable failure B34 Elastic-plastic cantilever plate under end moment B35 Cracking in a steel-lined concrete cylinder B36 Dynamic analysis of a plastic spherical cap B37 Removal and addition of elements, birth/death option B38 Analysis of temperatures and thermal stresses B39 Creep analysis of a cantilever B40 Analysis of snap-through of an arch structure B41 Snap-back of a bar structure B42 Nonlinear response of a shallow shell B43 Linearized buckling analysis of a stiffened plate B44 Stiffened plate with initial imperfections B45 Linearized buckling analysis of a circular arch Version 99.0 Nonlinear Examples 2

SOLVIA Verification Manual B46 Linearized buckling analysis of a column B47 Dynamic restart from static configuration B48 Cantilever subjected to deformation dependent loading B49 Plastic analysis of a thick walled cylinder B50 Thermal buckling of a rectangular plate B51 Large deflection analysis of a cable structure B52 Analysis of a cantilever subjected to end moment B53 Rectangular shaft subjected to twisting moment B54 Analysis with plastic-multilinear model, isotropic hardening B55 Analysis with plastic-multilinear model, kinematic hardening B56 Analysis of Hertz axisymmetric contact problem B57 Analysis of Hertz plane strain contact problem B58 Cable around a fast pulley B59 Dynamic analysis of frictional sliding of a point mass B60 Collision between a point mass and a rigid surface B61 Longitudinal impact of two identical bars B62 Analysis of Hertz dynamic contact problem B63 Analysis of elastic creep - creep law no. 1 B64 Analysis of elastic creep - creep law no. 2 B65 Large displacement analysis of a spherical shell B66 Analysis of one-dimensional seepage flow B67 Unconfined seepage flow through a rectangular dam B68 Unconfined seepage flow through a gravity dam B69 Semi-infinite region undergoing two phase changes B70 User-supplied material model (Ramberg-Osgood)

B71 Natural frequencies of a pre-loaded beam B72 Plastic collapse of a beam structure, pipe section B73 Plastic collapse of a beam structure, box-section B74 Analysis of a rubber bushing B75 Large deflection of a curved elastic cantilever B76 Initial strain in a curved beam B77 Temperature gradients in a curved beam B78 Temperature gradients in a plate B79 Reinforced concrete beam, ISOBEAM BS0 Reinforced concrete beam, nonlinear-elastic ISOBEAM B81 Bresler-Scordelis concrete BEAM B82 Compression of a rubber o-ring, PLANE STRAIN B83 Complex-harmonic analysis of a pre-loaded BEAM B84 Immersion of a shell structure into water, buoyancy effects B85 Pure shear of a rubber sheet B86 In-plane buckling of an unsymmetric deep arch B87 Lateral buckling of a right-angle frame B88 Lee's Frame Buckling Problem B89 Bresler-Scordelis concrete BEAM, A-2 with stirrups B90 Compound cylinder Version 99.0 Nonlinear Examples 3

SOLVIA Verification Manual B91 Concrete slab under uniaxial moment, Jain and Kennedy B92 Tilting-pad journal bearing, rigid pads B93 Tilting-pad journal bearing, elastic pads B94 Journal bearing, full Sommerfeld solution B95 Heat generation in a parallel-surface slider bearing B96 Tilting-pad journal bearing, elastic pads and shaft B97 Tilting-pad thrust bearing, spring supported B98 Sector-shaped thrust bearing, isoviscous oil and rigid pad B99 Simply supported concrete slab, Jain and Kennedy Version 99.0 Nonlinear Examples 4

SOLVIA Verification Manual EXAMPLE BI ELASTIC-PLASTIC ANALYSIS OF TRUSS STRUCTURE Objective To verify the geometric and material nonlinear behaviour of the TRUSS element.

Physical Problem A three-member truss structure is subjected to a concentrated load as shown in the figure below. The material of the structure is assumed to be elastic-perfectly-plastic. The variation of the applied load with time is also shown in the figure below. Py is the limit load considering small displacements only (Py = I00J2 + 300 = 441.42 kips) y Bar Areas = 1.0 ft2 E = 2.0.105 kips/ft2 ET = 0.0 a=5.0 ft (y = 100.0 kips/ft2 P. = 441.42 kips a

0 1

2 3

4 5

Time Finite Element Model In the finite element model considered, each member is represented by a 2-node elastic-plastic TRUSS element as shown in the top figure on page B 1.3. Large displacements are prescribed with stiffness reformation and BFGS equilibrium iterations at every step.

Version 99.0 P

a a

P/ P 1.01 0.00 Nonlinear Examples I

I 131.1

SOLVIA Verification Manual Solution Results The input data on page B 1.4 gives the results as shown in the table below. Due to symmetry, the force in element 1 in element group I is the same as in element 2.

Element group I Element group 2 Truss 1 Truss 1 Time Force [kips]

Force [kips]

1 26.12 52.23 2

52.23 104.45 3

78.35 156.66 4

100.00*

215.17 5

100.00*

300.00*

Plastic A check of equilibrium at time 5 gives, since the displacement in the Y-direction for node 4 is calculated to be 0.3276 14 inch:

Applied force : 1.01. 441.42 = 445.8342 kips Internal truss forces in the Y-direction:

5.327614 2.100.

+ 300 = 445.8342 kips

-r5 +5.3276142 The stress-strain solution points for the left and center elements as predicted in the finite element analysis are shown in the right bottom figure on page B 1.3. Note that the curves in the figures are drawn as straight lines between the solution points, which may not be the true stress-strain relation if, for example, there is a change from elastic to plastic condition. Deformed mesh and Y-displacement history for node 4 are shown in the left bottom figure on page B 1.3.

User Hints

" The option of large displacements must be used for a physically realistic modeling of the structure.

If a material-nonlinear-only analysis is employed, the structure is predicted to become unstable once the left and right elements become plastic since at that state there is no stiffness in the X translational degree-of-ffreedom at node 4. When large displacements are considered, the analysis proceeds beyond this load point, because the forces in all the members provide a geometric stiffness to the structural model.

" The TRUSS element is formulated for small strain conditions only. Therefore, the area of the TRUSS element is assumed to remain constant during the analysis and Poisson's ratio is not included in the input to the material models. Also, the strain calculated by SOLVIA for the TRUSS element in large displacement analysis is engineering strain and not Green-Lagrange strain. For example, the axial strain in element 2 at time 5 is u

0.327614 e= - =-0.06552 L

5.000000 Version 99.0 Nonlinear Examples B 1.2

SOLVIA Verification Manual Nonlinear Examples BI ELASTIC -P

,A STC ANALYSTS 3RTI7NAL j1 T 1,'E I TRUSS STRUCTURE Y Lxx 2-1

89. 1668 MASTER B EGNRIN SCLV1A ENGINEERTNG AB SCLVIA-PRE 99.0 B1 ELASTIC-PLAST:C ANALYSIS OF TRUSS STRLC ORIGINAL -

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,IM TIME Bt ELAST71-PLASTIC ANALYSIS OF TRUSS STRUCTURE 3,

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'Bi ELASTIC-PLASTIC ANALYSIS OF TRUSS STRUCTURE' DATABASE CREATE MASTER IDOF=001111 NSTEP=5 KINEMATICS DISPLACEMENT=LARGE TIMEFUNCTION 1

0.

/

5. 1.01 COORDINATES 1

-5.

TO 3

5.

/

4

0.

5.

MATERIAL 1

PLASTIC E=2.E5 YIELD=*00.

ET=0.

MATERIAL 2

PLASTIC E-2.E5 YIELD=300.

ET=O.

EGROUP 1

TRUSS MATERIAL=i ENODES

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1 1 4

/

2 3 4 EDATA

/

1

1.

EGROUP 2

TRUSS MATERIAL-2 ENODES

/

1 2 4 EDATA

/

1

1.

FIXBOUNDARIES

/

1 TO 3

LOADS CONCENTRATED 4 2 441.42 MESH VIEW=Z NNUMBERS=YES ENUMBERS=GROUP BCODE=ALL VECTOR=LOAD SOLVIA END SOLVIA-POST input Bi ELASTIC-PLASTIC ANALYSIS OF TRUSS STRUCTURE DATABASE CREATE WRITE FILENAME='bI.Iis' SET PLOTORIENTATION=PORTRAIT MESH VIEW=Z ORIGINAL=DASHED DMAX=1 VECTOR=REACTION SUBFRAME=12 NHISTORY NODE=4 DIRECTION=2 OUTPUT=ALL SYMBOL=I EXYPLOT ELEMENT=i POINT-i XKIND=ERR YKIND-SRR SYMBOL=1 SUBFRAME=i2 EGROUP 2

EXYPLOT ELEMENT=i POINT=1 XKIND=ERR YKIND=SRR SYMBOL=2 ELIST TSTART=i TEND=5 END Version 99.0 Nonlinear Examples BI1.4

SOLVIA Verification Manual EXAMPLE B2 LARGE AMPLITUDE OSCILLATION OF A PENDULUM Objective To verify the dynamical behaviour of the TRUSS element in large displacement analysis.

Physical Problem The simple pendulum shown in the figure below is released from a horizontal position at time t=0.

The response for one period of oscillation is to be determined.

Y L = 3.0443 m EA = 1.0- 10' N M = 10.0 kg g = 9.8 m/s 2

L M

Finite Element Model The pendulum is idealized as a TRUSS element capable of large displacements with a concentrated mass at its free end. The trapezoidal rule is employed to obtain the step-by-step response with a time step size of 0.1 second. Full-Newton iterations are performed at each step of the solution. The pa rameters used for measuring equilibrium iteration force convergence are RTOL=0.000 1 and RNORM=98.0 (pendulum weight) where RTOL is the force tolerance and RNORM is the reference load.

Solution Results The input data on pages B2.4 and B2.5 is used in the finite element analysis. The bottom figure on page B2.3 shows the time history of the angle 0, as defined in figure above, and the angular velocity 0, respectively, as calculated and displayed by SOLVIA-POST. The top figure on page B2.3 shows the time history of the axial force in the truss and the deformed mesh for time 1.0.

The maximum absolute value of the angular acceleration is predicted to be 3.219 rad/s2 which can be compared to the theoretical value g/L= 9.8/3.0443=3.219 rad/s 2 Version 99.0 Nonlinear Examples B2.1

SOLVIA Verification Manual The maximum absolute value of the angular velocity is predicted to be 2.50 rad/s (at time 3.1 sec) which can be compared to the theoretical value 2g/ L = V2.9.8 / 3.0443 = 2.54 rad/s The maximum axial force in the truss is predicted by SOLVIA to be 292.26 N which can be compared to the theoretical value 3 mg = 3.10-9.8 = 294.00 N User Hints

" To avoid spurious oscillations in the element force, it is important to select a small enough value of the Young's modulus times area for the truss material. In this example, the natural period for vibrations of the mass in the direction of the element is 2 TmU/BA = 0.1096 sec which is of the same order as the time step. An E-A -value of 1.0. o10'N, for example, would produce spurious axial force oscillations in the element if the time step is kept the same.

"* Note that the load vector input by time functions at the start of the solution (t=0) is not used in SOLVIA, see User Hints for Example A52. In this example, an initial acceleration is therefore pre scribed at time t=0.

"* SOLVIA-POST has the capability to evaluate a user-supplied arithmetic expression based on the nodal or element results. The results are calculated by SOLVIA referring to the global coordinates and a transformation to a cylindrical coordinate system can be done in SOLVIA-POST.

Version 99.0 Nonlinear Examples B2.2

SOLVIA Verification Manual Nonlinear Examples 82* -RGE A'IPL -7 7-CSCIL'ATLA'! OF A 7--N>LU OP:

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4.076 T-I1l I

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SOL / A-PCST 99.C 32 Z-ARGE AýIIPLITUDE OSCILLAT70N SQLVIA 7NGINEER:-NG AB OF A PENDUL'J>

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Version 99.0 B.

I 3

B2.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B2 LARGE AMPLITUDE OSCILLATION OF A PENDULUM' DATABASE CREATE MASTER IDOF=001111 NSTEP=45 DT=0.1 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK KINEMATICS DISPLACEMENTS=LARGE ITERATION METHOD-FULL-NEWTON TOLERANCES TYPE-F RTOL=1.E-4 RNORM=98.

COORDINATES 1

/

2 3.0443 MASSES

/

2

10.

INITIAL ACCELERATION

/

2

0.

-9.8 MATERIAL 1

ELASTIC E=1.E5 EGROUP 1

TRUSS ENODES

/

1 1 2 EDATA

/

1

1.

FIXBOUNDARIES

/

1 LOADS MASSPROPORTIONAL YFACTOR=1.

ACCGRA--9.8 SOLVIA END Version 99.0 Nonlinear Examples B2.4

SOLVIA Verification Manual SOLVIA-POST input B2 LARGE AMPLITUDE OSCILLATION OF A PENDULUM DATABASE CREATE TOLERANCES WRITE FILENAME='b2.1is' SUBFRAME 21 MESH VIEW=Z NNUMBERS=YES NSYMBOLS=YES ORIGINAL=YES TIME=1.

EHISTORY ELEMENT=I POINT=1 KIND=FR NPOINT NVARIABLE NVARIABLE NVARIABLE NVARIABLE NVARIABLE CONSTANT CONSTANT CONSTANT RESULTANT RESULTANT RESULTANT SUBFRAME RHISTORY RHISTORY RMAX RMAX RMAX EMAX END NAME=TWO NAME=DX NAME=VX NAME=VY NAME=AX NAME=AY NAME=DEG NAME=L NAME-PI NODE=2 DIRECTION=I DIRECTION=1 DIRECTION=2 DIRECTION=1 DIRECTION=2 180 3.0443 3.14159 KIND=VELOCITY KIND=VELOCITY KIND=ACCELERATION KIND=ACCELERATION NAME=THETA

'DEG/PI*ASIN((L+DX)/L)'

NAME=ANGVEL

'VX/L*COS(ASIN((L+DX)/L))+VY/L*(L+DX)/L' NAME=ANGACC

'AX/L*COS(ASIN((L+DX)/L))+AY/L*(L+DX)/LI 21 POINTNAME=TWO RESULTANTNAME=THETA POINTNAME=TWO RESULTANTNAME-ANGVEL ZONENAME=N2 ZONENAME=N2 ZONENAME=N2 RESULTANTNAME=THETA NUMBER=5 RESULTANTNAME=ANGVEL NUMBER=5 RESULTANTNAME=ANGACC NUMBER=5 Version 99.0 Nonlinear Examples B2.5

SOLVIA Verification Manual EXAMPLE B3 ELASTIC BUCKLING OF A CYLINDRICAL PANEL Objective To verify the large displacement behaviour of the PLATE element when subjected to a uniformly distributed loading.

Physical Problem A cylindrical panel clamped at all four edges, as shown in the figure below, is considered. The panel is subjected to a uniform pressure and the pre-and post-buckling behaviour of the panel is to be determined assuming that the material of the panel remains elastic.

a = 10.0 in h = 0.125 in

.1 R = 100.0 in T

C E = 4.5.10 5 psi h

a v0.3 q = 0.4 psi Y, V, Oy Finite Element Model Using symmetry considerations, only one-quarter of the panel is modeled. The finite element model, shown in the left figure on page B3.2, consists of a 4x4x4 mesh of PLATE elements. The maximum pressure load of 0.4 psi is reached in 20 equal load steps. Large displacements are prescribed in the analysis with stiffness reformation and BFGS equilibrium iterations performed at each load step.

Solution Results The relation of the central vertical deflection versus the applied loading, as predicted by the finite element model using the input data on page B3.3 is shown in the right figure on page B3.2. The solution of the finite element analysis is in good agreement with the results reported in [I].

Version 99.0 Nonlinear Examples B3.1

SOLVIA Verification Manual Nonlinear Examples User Hints

"* Note that the program treats this instability analysis as a large deflection problem and does not indicate a bifurcation point. The bifurcation would be indicated by a zero determinant, which in practice means a very small positive or negative pivot, due to round-off.

"* This analysis can also be performed very effectively using the automatic load step incrementation method in SOLVIA.

Reference

[1]

Dhatt, G., "Instability of Thin Shells by the Finite Element Method", IASS Symp. for Folded Plates and Prismatic Structures, Vienna, 1970.

33 ELAST C BUJKLANG OF A C'"LANCRRCL PANEL OR GINAL 2

S D

E ORIGINAL Z

xA1--, y MAS TR 000050 OtolOl IA X

y B3 ELASTIC BUCKLING OF A CYLANDRO AL PANEL ORIGINAL

2.

MAX D7SPL.

0.39758 T0rE 20 4t

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97. 397 Z-DIR D7SPL.

MAX 0 2.074347

"-. :2424

-0. 17394

'-0. 22364

-0. 27334

-0. 32304 j-0.37273

'IN-0 139758 SCLUIA-RE 99 0

B3.2 SOLV[A

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B3 ELASTIC BUCKLING OF A CYLINDRICAL PANEL' DATABASE CREATE MASTER NSTEP-20 KINEMATICS DISPLACEMENT=LARGE TIMEFUNCTION 1

0.

0. /

20.

0.4 SYSTEM 1 CYLINDRIC COORDINATES ENTRIES NODE R

THETA XL 1

100.

90.

0.

2 100.

90.

10.

3 100.

84.27042 10.

4 100.

84.27042

0.

MATERIAL 1

ELASTIC E=4.5E5 NU=0.3 EGROUP 1

PLATE GSURFACE 1 2 3 4 EL1=4 EL2=4 SYSTEM=1 EDATA

/

1 0.125 LOADS ELEMENT INPUT=ELEMENTS 1 T 1.

TO 64 T 1.

FIXBOUNDARIES 123456 INPUT:LINE

/

2 3

/

3 4 FIXBOUNDARIES 246 INPUT-LINE

/

1 2 FIXBOUNDARIES 156 INPUT LINE

/

4 1 SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES MESH OUTLINE=YES NNUMBERS=MYNODES BCODE=ALL SUBFRAME=12 MESH VECTOR=LOAD SOLVIA END SOLVIA-POST input B3 ELASTIC BUCKLING OF A CYLINDRICAL PANEL DATABASE CREATE WRITE FILENAME='b3.1is' SET PLOTORIENTATION=PORTRAIT OUTLINE=YES MESH DMAX=1 CONTOUR=DZ ORIGINAL=YES VECTOR=REACTION SUBFRAME=12 NHISTORY NODE=1 DIRECTION=3 XVARIABLE=1 OUTPUT=ALL SYMBOL=i END Version 99.0 Nonlinear Examples B3.3

SOLVIA Verification Manual EXAMPLE B4 ELASTIC BUCKLING OF AN IMPERFECT SQUARE PLATE Objective To verify the large displacement behaviour of the cubic SHELL element when employed in buckling analysis.

Physical Problem The square plate considered in the analysis is shown in the figure below. The plate is simply supported on all edges and uniform compressive loads are applied at the two opposite ends.

Thickness h

b=2.0 in h = 0.01 in E= 1.0.108 psi v=0.3 p

p (psi) 9040-0 2

4 6

6 10 Time Finite Element Model The finite element model is shown in the left figure on page 132.2. Using symmetry conditions, only one-quarter of the plate is modeled with one cubic SHELL element. The following initial imperfec tion in geometry is imposed on the plate:

zl(XY)=

ht 2X 1 2YI The above equation represents a linear approximation to the first buckling mode shape of the perfect plate. The loads are applied in ten equal load steps and a geometrically nonlinear analysis (option DISPLACEMENTS = LARGE) with stiffness reformation and BFGS equilibrium iterations in each step is performed.

Version 99.0 P

b Nonlinear Examples 134.1

SOLVIA Verification Manual Nonlinear Examples Solution Results The displacement at the center of the plate versus the applied loading, as predicted by the finite element model using the input data on page B4.3, is shown in the right bottom figure on page B4.2.

The analytical buckling load for the perfect plate using [1]. p. 352 is given by p, = 9038 psi The deformed element mesh is also shown in the right figure on page B4.2.

User Hints

"* The shape of the initial imperfection in this problem is assumed to be known. In general, the initial imperfection geometry can be automatically generated by SOLVIA as described in Example B44.

"* Note that a lower integration order than the default 4x4x2 Gauss integration can produce spurious kinematic modes in the element and, hence, unreliable results.

Reference

[1]

Timoshenko, S.P., and Gere, J.M., Theory of Elastic Stability, 2nd Edition, McGraw-Hill, 1961.

84 OLAST:C AUCKLING OF AN IMPERFECT SOUARE PL ORIGINAL -

0.2 F

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I 0 3 cC

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MAX D SPL.

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Y LOAD 33.901 Z-DIR DISPL MAX 2,7678E-3 CAT2.$948E3 S2.2188E-3 i:!, 9029E-3 St.Sý69E-3 S1.21ý9E-31

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B4 ELASTIC BUCKLING OF AN IMPERFECT SQUARE PLATE' DATABASE CREATE MASTER NSTEP=10 KINEMATICS DISPLACEMENTS=LARGE TIMEFUNCTION 1

0.

/

10.

9040.

COORDINATES 1

0.

1.E-4

/

2

1.

/

3

1.

/

4

0.

1.

MATERIAL 1

ELASTIC E=1.E8 NU=0.3 EGROUP 1

SHELL GSURFACE 1 2 3 4 EL1=1 EL2-1 NODES=16 THICKNESS 1

0.01 LOADS ELEMENT TYPE=FORCE INPUT=LINE 2 3 out -0.01 -0.01 FIXBOUNDARIES 36 INPUT=LINE 3

4

/

2 3 FIXBOUNDARIES 156 INPUT=LINE

/

4 1 FIXBOUNDARIES 246 INPUT=LINE

/

1 2 SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES MESH NNUMBERS=MYNODES BCODE=ALL SUBFRAME=12 MESH EAXES=RST VECTOR=LOAD SOLVIA END SOLVIA-POST input B4 ELASTIC BUCKLING OF AN IMPERFECT SQUARE PLATE DATABASE CREATE WRITE FILENAME='b4.1is' SET PLOTORIENTATION=PORTRAIT ORIGINAL=YES MESH CONTOUR=DZ NSYMBOLS YES DMAXN1 VECTOR=LOAD SUBFRAME=12 NHISTORY NODE=i DIRECTION=3 SYMBOL=i XVARIABLE=1 OUTPUT=ALL END Version 99.0 Nonlinear Examples B4.3

SOLVIA Verification Manual EXAMPLE B5 CANTILEVER BEAM IN CREEP Objective To verify the PLANE STRESS2 element for material-nonlinear-only analysis using the creep material model.

Physical Problem The cantilever beam, shown in the figure below, is subjected to a constant bending moment applied at the tip. The material of the beam is assumed to obey the uniaxial creep law

,= 6.4."10 '

.15"t in/in where 7 is measured in psi and t in hours.

z T

NELT A

t--

Y h-b = 0.3 in E = 3.0.10 7 psi h=4.0in v=0.3 L = 40.0 in 5c = 6.4. 10 -A -0315 in /in /hr M = 6000.0 lb-in Finite Element Model The finite element model considered for the analysis consists of eight 8-node PLANE STRESS2 elements as shown in the figure on page B5.2. By restraining the Y-displacement at the neutral axis, only the portion of the beam above the neutral axis is modeled. For the analysis, a step size of ten hours and twenty time steps with BFGS equilibrium iterations in each step is used. For the integration of the creep response the Euler backward method is used, i.e. the parameter a. is set to 1.0, and one subdivision per time step is employed. The finite element stiffness matrices are evaluated using 3x3 Gauss integration.

Solution Results The input data on pages B5.3 and B5.4 are used in the finite element analysis. The figure on page B5.3 shows the time history response of the bending stress at element 6, integration point 1 (Y = 18.873, Z = 1.887). The theoretical steady-state value at the same location is 21'+a 1+1 M 2a, +l(h2)

Z -=5688psi 2'

2b a,

Version 99.0 Nonlinear Examples B5.1

SOLVIA Verification Manual Nonlinear Examples The figure on page B5.3 shows the deformed mesh at the time t=200 hours.

User Hints For materially nonlinear analysis it may be important to use higher order Gauss integration and more elements than in linear analysis. In a linear analysis only one 8-node PLANE STRESS element through the thickness of the beam would yield excellent results.

A variable time step solution of this problem is performed in Example B39.

B5 CANTILEVER BEAM

N CREEP OR--GINAL S.

Z L.

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. S TIME 1C 1

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N C, ORKGINAL MAX D:SPL TINE 200 LOAD 1125 F

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1. /

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2 3

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0. 0.

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INITIAL TEMPERATURES TREF=200.

MATERIAL 1 PLASTIC-CREEP ISOTROPIC XKCRP 1 TREF=200.

ALPHA=1, XISUBM-1 A0=6.4E-18 A1=3.15 A2=1.

0.

3.E7 0.3 3.E7 3.E6 0.

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EGROUP 1

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2 3 4 ELi=4 EL2=2 NODES=8 EDATA

/

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3 0.

-7500.

FIXBOUNDARIES 2 INPUT=LINES

/

1 2 FIXBOUNDARIES 3 INPUT=NODES

/

1 LOADS TEMPERATURES INPUT=SURFACE 1 2 3 4 200.

1 4 SUBFRAME 12 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL MESH EAXES=RST VECTOR=LOAD SOLVIA END SOLVIA-POST input B5 CANTILEVER BEAM IN CREEP DATABASE CREATE WRITE FILENAME='b5.1is' SUBFRAME 12 MESH ORIGINAL=DASHED SMOOTHNESS=YES VECTOR=LOAD EHISTORY ELEMENT=6 POINT=i KIND-SYY SYMBOL=i OUTPUT=ALL, SUBFRAME=2121 NHISTORY NODE=2 DIRECTION=3 SYMBOL=i OUTPUT=ALL SUBFRAME=2221 NLIST END KIND=REACTION Version 99.0 Nonlinear Examples B5.4

SOLVIA Verification Manual EXAMPLE B6 PIPE WHIP ANALYSIS, DIRECT INTEGRATION SOLUTION Objective To demonstrate the dynamic behaviour of the BEAM and TRUSS elements in elastic-plastic analysis with gap.

Physical Problem A cantilever pipe, shown in the figure below, is initially at rest and is suddenly subjected to a constant tip load, causing it to impinge onto a restraint.

0E-

-T-Do diameter d restraint D0 = 30.0 in Di = 27.75 in L = 360.0 in a= 3.0 in b=21.0 in d = 5.75 in a

C"

-.125

/E F

Pipe material:

E = 2.698.10 7 psi v=0.3 a, = 2.914. 104 psi p = 7.18-_10-4 lbs!(g

in3 )

Restraint material:

E = 2.99.10 7 psi Gy = 3.80. 104 psi Version 99.0 Nonlinear Examples t

B6.1

SOLVIA Verification Manual Finite Element Model The pipe is modeled using six BEAM elements and the pipe material is assumed to be elastic perfectly-plastic. The restraint is represented by a TRUSS element with elastic-perfectly-plastic material and a gap width of 3 in. The finite element model used for the analysis is shown in the top figure on page B6.3. The Newmark method is used for the step-by-step direct integration and the response is calculated for one thousand time steps using a time step of 0.0001 sec. Full-Newton iterations with line-search are performed in every step using an energy tolerance of 1.0.10-8. A lumped mass matrix assumption is employed in the analysis. The BEAM element stiffness matrices are evaluated using 5xlx5 Newton-Cotes integration.

Solution Results The tip and mid node responses are shown on pages B6.3 and B6.4. The axial stress in the TRUSS element as a function of time and the variation of the axial stress in element I of the pipe are shown in figure on page B6.5. The SOLVIA numerical solution is obtained by using the input data on pages B6.5 and B6.6.

User Hints

"* The characteristics of this problem are that the pipe is very flexible and impinges suddenly onto a relatively stiff restraint. The nonlinearity is due to the gap and the plasticity in the restraint and in the pipe.

"* The restraint can also be modeled using a nonlinear elastic material model and the TRUSS element without the gap option if only the first portion of the response (without unloading from the plastic state) is of interest.

Version 99.0 Nonlinear Examples B6.2

SOLVIA Verification Manual Nonlinear Examples B6 PTDI IUH7P ANAL CRIGNAL T IME

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'B6 PIPE WHIP ANALYSIS, DIRECT INTEGRATION SOLUTION, DATABASE CREATE MASTER IDOF=001110 NSTEP=1000 DT=1.E-4 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=-.E-8 COORDINATES 1

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360.

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MATERIAL MATERIAL I PLASTIC E=2.698E7 NU=0.3 YIELD=29140 ET=0.

DEN=7.18E-4 2

PLASTIC E=2.990E7 NU=0.

YIELD=38000 ET=0.

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EGROUP 2

TRUSS MATERIAL=2 GAP YES ENODES

/

1 8 7 EDATA

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ENTRIES EL AREA GAPWIDTH 1

25.967 3.

FIXBOUNDARIES

/

1 8 LOADS CONCENTRATED 7 2 -6.57E5 SET MESH MESH NSYMBOLS=YES VIEW=Z NNUMBERS=YES VECTOR=LOAD SUBFRAME=12 ENUMBERS=YES BCODE=ALL SOLVIA END SOLVIA-POST input B6 PIPE WHIP ANALYSIS, DIRECT INTEGRATION SOLUTION DATABASE CREATE WRITE FILENAME='b6.1is' SUBFRAME NHISTORY NHISTORY SUBFRAME NHISTORY NHISTORY SUBFRAME NHISTORY NHISTORY 21 NODE=7 NODE=7 21 NODE=7 NODE=4 21 NODE-4 NODE=4 DIRECTION=2 KIND=DISPLACEMENT DIRECTION=2 KIND=VELOCITY DIRECTION=2 KIND=ACCELERATION DIRECTION=2 KIND=DISPLACEMENT DIRECTION:2 KIND=VELOCITY DIRECTION=2 KIND=ACCELERATION SUBFRAME 21 EHISTORY ELEMENT=-

POINT=515 KIND=SRR EGROUP 2

EHISTORY ELEMENT=i POINT=1 KIND=SRR NMAX DIRECTION=2 END Version 99.0 Nonlinear Examples B6.6

SOLVIA Verification Manual EXAMPLE B7 PIPE WHIP ANALYSIS, MODE SUPERPOSITION Objective To verify the use of mode superposition in nonlinear analysis.

Physical Problem A cantilever pipe, shown in the figure below, is initially at rest and is suddenly subjected to a constant tip load, causing it to impinge onto a restraint. The same physical problem is already analyzed in Example B6 using direct integration which serves as a reference solution.

P

-tI 6 -57 x106 'b a

diameter d b

restraint D0 = 30.0 in Di = 27.75 in L = 360.0 in a= 3.0 in b=21.0 in d = 5.75 in a-T-

-.125

- My Pipe material:

E =2.698.107 psi v=0.3 (y, = 2.914.104 psi Restraint material:

E = 2.99-10 7 psi CTy = 3.80-104 psi p = 7.18.10-4 lbs/(g, in3 )

Finite Element Model The same model as in Example B6 is used, see the top figure of page B7.3. The frequencies and mode shapes are calculated based on the initial configuration using the subspace iteration method. The mode shapes are employed in the nonlinear time integration using the modified Newton method with the energy tolerance 1.0. 108.

Version 99.0 Nonlinear Examples B7.1

SOLVIA Verification Manual The model has 6 nodes with translational lumped masses in two directions. The 6 lowest bending mode shapes are of interest here. The axial mode shapes give no contribution to the response since the load acts in the transverse direction. The first run is based on 8 modes, which include all the 6 bend ing modes with frequencies from 8.4 Hz to 553 Hz. The time step is then selected to 10-" sec which gives about 18 steps per highest period.

The second run with 2 modes was performed with the time step 0.25.10-3 sec giving about 80 steps per highest period. The time step is shorter than in a conventional linear modal superposition because of the nonlinearities due to impact and plasticity.

Solution Results The input data on pages B7.8 and B7.9 is used in the finite element analysis with 8 modes. The shapes of the first 2 modes are shown in the bottom figure on page B7.3.

The tip and mid node responses (displacements, velocities, accelerations) predicted in the analysis are shown in the figures on pages B7.4 and B7.5. The axial stress in the TRUSS element as a function of the axial strain and the variation of the axial stress in element 1 of the pipe are also shown in the bottom figures on page B7.5.

We note that almost the same response as in the reference solution given in Example B6 is obtained.

It is not possible to obtain exactly the same solution using modal superposition with a constant set of mode shapes. The actual mode shapes change somewhat during the solution since the stiffness varies due to development of plasticity.

The solution with 2 modes is approximate, see pages B7.6 and B7.7. It displays, however, the main characteristics of the response. The tip response of the pipe and the axial stress response in the restraint are in good agreement with the reference solution.

User Hints

"* The mode superposition solution is very effective when only a few modes need to be used to represent the response to sufficient accuracy. The pipe whip example solved here is such a case.

" Note that the mode shapes of the linear structure (corresponding to time 0) are used in the mode superposition solution, The mode shapes satisfy

'K p = wo M P When 2 modes are employed the response is calculated using 2

U =

ai (tjpi where c. (t) are time dependent coefficients evaluated in the solution. In essence, therefore, a two degree-of-freedom finite element model is considered with the displacement assumptions p, and ýp2 but all elastic-plastic nonlinearities of the pipe and the restraint based on the displacement assumption are included in the analysis.

"* In nonlinear mode superposition analysis only modified Newton equilibrium iteration is available.

Version 99.0 Nonlinear Examples B7.2

SOLVIA Verification Manual 57 P7PE *:-ZP ANA 0RTI GNA so,4 2

IS, MODE SUPcRPOS!T!CN, 8 MODES 5

ORIGNAL i

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SOLVIA Verification Manual B7 PIPE WHIP ANALYSIS, MODE SUPERPOSJTION, 8 MCDES C

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SOLVIA Verification Manual Nonlinear Examples 37 PIPE

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ANALYSIS, MODE SUPE PCSI

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'B7 PIPE WHIP ANALYSIS, MODE SUPERPOSITION, 8 MODES' DATABASE CREATE MASTER IDOFO001110 NSTEP=1000 DT=1.E-4 FREQUENCIES SUBSPACE-ITERATION NEIG=12 SSTOL=1.E-10 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED IMODS=-

NMODES=8 ITERATION METHOD=MODIFIED-NEWTON TOLERANCES ETOL=lE-8 COORDINATES 1

TO 7

360.

/

8 360.

-24.

MATERIAL 1

PLASTIC E=2.698E7 NU=0.3 YIELD=29140 DENSITY=7.18E-4 MATERIAL 2

PLASTIC E=2.990E7 NU=0.

YIELD=38000 ET=0.

DEN=0.

EGROUP 1

BEAM DIM2 RINT=5 SINT=1 TINT=5 MATERIAL=i RESULT=TABLE STRESSTABLE 1

11 115 411 415 511 515 SECTION 1

DIMENSION PIPE DO=30.

DI=27.75 ISHEAR=1 ENODES

/

1 8 1 2

TO 6

8 6 7 EGROUP 2

TRUSS MATERIAL 2 GAP=YES ENODES

/

1 8 7 EDATA

/

ENTRIES EL AREA GAPWIDTH 1

25.967 3.

FIXBOUNDARIES

/

1 8 LOADS CONCENTRATED 7 2 -6.57E5 MESH VIEW=Z NSYMBOLS=YES NNUMBERS=YES VECTOR=LOAD SUBFRAME=12 MESH VIEW=Z NSYMBOLS=YES BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B7.8

SOLVIA Verification Manual SOLVIA-POST input B7 PIPE WHIP ANALYSIS, MODE SUPERPOSITION, 8 MODES DATABASE CREATE WRITE FILENAME=-b7.1is' FREQUENCIES SET RESPONSETYPE=VIBRATION ORIGINAL=YES VIEW=Z MESH NSYMBOL=YES TIME=i SUBFRAME=12 MESH NSYMBOL=YES TIME=2 21 NODE=7 DIRECTION=2 NODE=7 DIRECTION=2 21 NODE=7 DIRECTION=2 NODF=4 DIRECTION=2 21 NODE=4 DIRECTION=2 NODE=4 DIRECTION=2 KIND=DISPLACEMENT KIND=VELOCITY KIND=ACCELERATION KIND=DISPLACEMENT KIND=VELOCITY KIND=ACCELERATION SUBFRAME 21 EHISTORY ELEMENT=i POINT=515 KIND=SRR EGROUP 2

EHISTORY ELEMENT=1 POINT=i KIND=SRR END Version 99.0 SUBFRAME NHISTORY NHISTORY SUBFRAME NHISTORY NHISTORY SUBFRAME NHISTORY NHISTORY Nonlinear Examples B7.9

SOLVIA Verification Manual EXAMPLE B8 CYCLIC LOADING OF PLASTIC TRUSS, ISOTROPIC HARDENING Objective To verify the behaviour of the 3-node TRUSS element in MNO (Mlaterially-Nonlinear-Only) analysis using the plastic isotropic hardening material model.

Physical Problem A bar subjected to an axial force as shown in the figure below is considered. The material of the bar is assumed to obey an elastic-plastic material law with isotropic hardening.

P L

R A A= 0.01m' L = 1.0 m E =2.0-110" N/m2 Gy = 4.0.108 N/m 2 ET =2.0-10'° N/m 2 E

The finite element model consists of one 3-node TRUSS element and the element stiffness matrix is evaluated using 2-point Gauss integration. The applied force is cyclicly varied to obtain plastic response both in tension and compression of the truss. Stiffness reformation and BFGS equilibrium iterations are performed at every time step.

Solution Results The input data on page B8.3 is used in the finite element analysis. The figure on page B8.2 shows the stress-strain response in the TRUSS element predicted in the analysis as displayed by SOLVIA POST. A hand calculation shows that the response predicted by SOLVIA agrees with the analytical solution of the model.

Version 99.0 Gy Finite Element Model Nonlinear Examples B8.1

SOLVIA Verification Manual Nonlinear Examples User Hints:

"* To model this example, of course, only a 2-node truss and 1-point Gauss integration need be employed. The results for the 2 Gauss points used are exactly the same and the midnode displacement is exactly half of the displacement at the end node.

" It is important to use the BFGS or the full Newton method of equilibrium iteration in this example.

The modified Newton method of iteration does not converge at unloading in this example from plastic to elastic conditions, and has difficulty to converge when loading from elastic to plastic conditions.

B8 JVCLIC LCADTNG 4r CK L0O a_

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'B8 CYCLIC LOADING OF PLASTIC TRUSS, ISOTROPIC HARDENING, DATABASE CREATE MASTER IDOF=l01111 NSTEP=9 TIMEFUNCTION 1

0.

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1.

2.0 /

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MATERIAL 1

PLASTIC ISOTROPIC E=2.E11 YIELD=4.E8 ET=2.E10 EGROUP 1

TRUSS ENODES

/

1 1

3 2 EDATA

/

1 0.01 FIXBOUNDARIES

/

1 LOADS CONCENTRATED 3 2 1.E6 SOLVIA END SOLVIA-POST input B8 CYLIC LOADING OF PLASTIC TRUSS, ISOTROPIC HARDENING DATABASE CREATE WRITE FILENAME='b8.1is' SUBFRAME 21 EXYPLOT EL=i POINT=i XKIND-ERR YKIND=SRR OUTPUT=ALL SYMBOL=1 NHISTORY NODE=3 DIRECTION=2 XVARIABLE=1 OUTPUT=ALL SYMBOL:1 END Version 99.0 138.3)

SOLVIA Verification Manual Nonlinear Examples EXAMPLE B9 CYCLIC LOADING OF PLASTIC TRUSS, KINEMATIC HARDENING Objective To verify the behaviour of the 3-node TRUSS element in MNO (Materially-Nonlinear-Only) analysis using the plastic kinematic hardening material model.

Physical Problem Same as in Example B8.

Finite Element Model Same as in Example B8 except that the material of the TRUSS element assumes an elastic-plastic kinematic hardening material law and other time function values.

Solution Results The input data on pages B9.2 is used in the finite element analysis. The figure below shows the stress strain response at one integration point of the TRUSS element during the load history in the finite element analysis as displayed by SOLVIA-POST. A hand calculation shows that the response predicted by SOLVIA agrees with the analytical solution of the model.

B9 CY7LIC LCADING OF PLASTIC I..........

TRUSS, 0

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0.

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PLASTIC KINEMATIC E=2.E11 YIELD=4.E8 ET=2.E10 EGROUP 1

TRUSS ENODES

/

1 1 3 2 EDATA

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/

1 LOADS CONCENTRATED 3 2 1.E6 SOLVIA END SOLVIA-POST input B9 CYLIC LOADING OF PLASTIC TRUSS, KINEMATIC HARDENING DATABASE CREATE WRITE FILENAME='b9.1is' SUBFRAME 21 EXYPLOT EL=1 POINT=1 XKIND-ERR YKIND=SRR OUTPUT=ALL SYMBOL=i NHISTORY NODE=3 DIRECTION=2 XVARIABLE=1 OUTPUT=ALL SYMBOL=1 END Version 99.0 Nonlinear Examples 139.2

SOLVIA Verification Manual EXAMPLE B1O CABLE UNDER GRAVITY LOAD Objective To verify the large displacement behaviour of the TRUSS element and demonstrate the modeling of analyses of cables.

Physical Problem A cable stretched between a ground anchor point and a tower attach point, see figure below, is considered. The cable has a cluster of 6 insulators attached to a point in the middle portion and 3 insulators attached to the lower portion. The weight of each insulator is 510 lbs. The initial prestress of the cable is 7520 lbs and the self weight is 0.106667 lb/in.

Tower attach point Area A

A= 0.361 in 2 L, = 8191.20 in Infctprestress 7520.0 Lb L, =7312.05 in E = 1.9.10 7 psi L, = 0.295476 lb/in 3 yIA=0.106667 lb/in Insulator y

Anchor point i'

Ly Finite Element Model The figure on page B 10.3 shows the finite element model. It consists of twelve 2-node TRUSS elements. The insulators are attached as concentrated masses at the nodes 2, 4, 6 and 8. The material model for the TRUSS elements is linear elastic and a large displacement analysis is prescribed.

Version 99.0 Nonlinear Examp les B 10.1

SOLVIA Verification Manual Solution Results The input to SOLVIA-PRE and SOLVIA-POST are shown on pages B 10.5 and B 10.6. The Full Newton method of equilibrium iteration is used. The energy tolerance is set to ETOL = 1.0. 10-5.

The gravity loading is applied in 5 equal load steps. The calculated deformed configuration and the displacements for node 8 in the Y-and Z-direction are given in figures on page B 10.4. We note that the cable stiffens as the gravity load is increased. For analytical verification we select the following results as typical:

Element Truss Internal Force (lbs) 7 22843.5 8

24975.0 Using the above solution the following internal TRUSS element forces can be calculated theoretically:

Element 7:

F7 = FP +

L7 EA = 22844 (lbs) whL7e where Fp = initial prestress, 17 =final length, L7 = original length, Element 8:

8=I

+i I EA =24975 (lbs)

The fifth digit is uncertain due to the number of digits in the listed results. Considering this accuracy it is concluded that the calculated values for F7 and F3 agree with the corresponding SOLVIA results.

By projecting the SOLVIA calculated internal forces for elements 7 and 8 onto the vertical (z) axis it can be seen that equilibrium is satisfied in the vertical direction for node 8.

Version 99.0 Node Displacements (inch)

Deformed Coordinates (inch) y z

y z

7 215.746

-248.196 3421.8026 2613.7601 8

242.152

-275.898 4132.1693 3196.6091 9

199.365

-226.344 4949.6183 4014.0711 Nonlinear Examples B 10.2

SOLVIA Verification Manual User Hints

" The initial prestress in the TRUSS elements is important since it provides stiffness in the lateral direction for the elements in a large displacement analysis. Without the prestress the stiffness matrix in this example would be singular and no solution would be possible. See also Example B5 1.

" The steady-state solution for gravity loading is considered a static load case in SOLVIA. The lumped masses of the model are assembled and only used to calculate the corresponding concen trated weight forces (using the acceleration of gravity).

The internal forces and the stresses in the TRUSS elements are calculated using the small strain assumption. but large displacements and rotations are included in the large displacement formulation.

Hence, here the area of a TRUSS element remains constant and the element strain is equal to (final length) - (original length)

(original length)

"* In a stiffening problem it may be necessary to use force tolerances in combination with the energy tolerance during the equilibrium iterations. Small variations in displacement increments may give relatively large variations in unbalance forces.

3'.0 CABLE UNDER GRAVITY LOAD ORzGINAL

.OO.

Z ORTGINAL 1000 Z

-0

.L y

1000y Y1 Y*

"6 3

I 43ll 02/IZA ENGNEERING AB SOLVIA-PRE 99.0 Version 99.0 Nonlinear Examples B 10.3

SOLVIA Verification Manual 3'0 QABLE UNDER GRAVITY LOAD FORCE-R NO AVERAGING MAX 25325.3 S25093.8 21630.8 243O 8 24t67.8 S23704.8 2324. 8 22778.8 S223'5. 8 21852.8 MIN 20621.3 3LVIA ENGINEERINO A3 SOLVIA-POST 99.0 810 CABLE UNDER GRAVITY

- I C!)

! L 7

c))

01-FTIF O VIA-ST 99.0 YIME SCL',/!.A NO-INEERING A Version 99.0 MAX D:SP.

TI"E S

--- 367.09 Nonlinear Examples

-CAD B 10.4

SOLVIA Verification Manual SOLVIA-PRE input HEAD

'Bi0 CABLE UNDER GRAVITY LOAD' DATABASE CREATE MASTER IDOFlI00111 NSTEP=5 ANALYSIS TYPE=STATIC MASSMATRIX=LUMPED KINEMATICS DISPLACEMENT=LARGE TOLERANCES EF ETOL=1.E-5 RTOL=1.E-2 RNORM=500.

ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TIMEFUNCTION 1

0. 0. /

5.

1.

COORDINATES ENTRIES NODE Y

Z 1

0.

2 306.049 273.201 3

977.988 873.022 4

1649.927 1472.843 5

2086.012 1862.123 6

2522.097 2251.404 7

3206.057 2861.956 8

3890.017 3472.507 TO 13 8191.200 7312.050 MASSES 2

0.

510.

4

0.

510.

6

0.

510.

8

0.

3060.

MATERIAL 1

ELASTIC E=1.9E7 DENSITY=0.295476 EGROUP 1

TRUSS ENODES

/

1 1 2 TO 12 12 13 EDATA ENTRIES EL AREA INIT-STRAIN 1

0.361 1.09637E-3 TO 12 0.361 1.09637E-3 FIXBOUNDARIES

/

1 13 LOADS MASSPROPORTIONAL ZFACTOR--1.

ACCGRA=1.

MESH VIEW=X NSYMBOLS=YES NNUMBERS=YES SUBFRAME=21 MESH VIEW=X NSYMBOLS=YES ENUMBERS YES BCODE=ALL GSCALE=OLD SOLVIA END Version 99.0 Nonlinear Examples B 10.5

SOLVIA Verification Manual SOLVIA-POST input B10 CABLE UNDER GRAVITY LOAD DATABASE CREATE WRITE FILENAME='bi0.1is' MASS-PROPERTIES SET VIEW=X CONTOUR AVERAGE=NO MESH NSYMBOLS=YES CONTOUR=FR MESH ORIGINAL=YES DEFORMED=NO TEXT=NO AXES=NO GSCALE-OLD, NSYMBOLS=YES SUBFRAME=OLD SUBFRAME NHISTORY NHISTORY NLIST ELIST GLIST SUMMATION SUMMATION END 21 NODE=8 DIRECTION=2 SYMBOL=i NODE=8 DIRECTION=3 SYMBOL=i KIND=LOAD KIND=REACTION DETAILS=YES Version 99.0 Nonlinear Examples B 10.6

SOLVIA Verification Manual EXAMPLE Bll RUBBER SHEET IN LARGE STRAIN ANALYSIS Objective To verify the large displacement and large strain behaviour of the PLANE STRESS2 element employing the rubber model.

Physical Problem A rubber sheet subjected to a uniform end load, see figure below, is considered. The material is assumed to be of the Mooney-Rivlin type, for which experiments by Iding [1, 2] gave C1 =21.605 lb.!in.2, C2 =15.747 lb./in.2 8.92 in P

THICKNESS =0.125 in RUBBER SHEET Finite Element Model By symmetry only one half of the rubber strip need be modeled, see figure on page B 11.2. Thirtythree 4-node PLANE STRESS2 elements are used and the end loading is applied as concentrated forces.

Solution Results The final total end load of 41.80 lb. is reached in 4 equal load steps using modified Newton iteration and stiffness reformation for every step. The tolerance is ETOL= 1.E-5. The input data used are shown on pages B 11.4 and B11.5.

Note that the scale for the displacements is the same as for the geometry. The end displacements corresponding to the load steps are shown in the top figure on page B 11.3 and good agreement with experimental results by Iding is obtained. A contour plot of the maximum principal stretch and a vector plot of the principal stretches in the YZ-plane are also shown in the figure.

The stress in the Y-direction is shown as a function of the Green-Lagrange strain for element 1I, integration point 1 in the left bottom figure on page B 11.3. The stress distributions at sections A-A and B-B are shown in the right bottom figure on page B 11.3, respectively. These stress distributions are also shown as displayed by SOLVIA-POST.

Version 99.0 Nonlinear Examples B11. 1

SOLVIA Verification Manual User Hints

"* The modified Newton iteration method is quite economical to use in this example, since nonlinearities monotonically soften the structure. However, the nonlinearities are very significant.

"* Using 8-node PLANE STRESS2 elements much less elements could be used in the model for an accurate response prediction.

References

[1]

Iding, R.H., "Identifications of Nonlinear Material by Finite Element Methods", Report UC SESM 73-4, Department of Civil Engineering, University of California, Berkeley, Jan., 1973.

[2]

"NONSAP - A Structural Analysis Program for Static and Dynamic Response of Nonlinear Systems", Report UC SESM 74-3, Department of Civil Engineering, University of California, Berkeley, February 1974.

B11 RUBBER SHEET IN LARGE STRAIN ANALYSIS ORIG:NAL C.-

ORIGINAL L

I.

7 N AS r

1001:

B to I l C I I I,

z MAX 45. 838 S42.974 37 244 S..

3 t. S 4

2S...

5 784 20 0S4 1i4.325 8.594 7 S2.8649 "1:N 0 SOLVIA ENýGINEERING AB SOLV/A-PRE 99.0 Version 99.0 Nonlinear Examples Bl11.2

SOLVIA Verification Manual Nonlinear Examples B11 RUBBER SHEET IN LARGE STRAIN ANALYSIS oRIG:NAL,-

2.,

MAX D SPL

8. 98' 7 XEE 4

'N MAX DGSPL 8.98'7 TIME 4

~STRET>4 I i

2. 155 0.68277 SOLV!A ENGINEERING AB

OLVIA-POST 99.0 TIME t

I Ar SOLVCA ENGINEERING AB Version 99.0 PINC IPAL y STRET H

,AX "1AX 2.1498 2 ý 88.

1.9647 1.84:.3 1 7179 1

594,

i 4711 MTN 1. 1621 z

Btl RUBBER SHEET IN LARGE STRAIN ANALYSIS I' E 2

2 3,

3.5

1.

2A SSVIA I

AS L

55 A

SOLVIA-POST 99.0 SOLVTA ANIERN B

SOL/TA-PO5T 99 3 P_,

Bl11.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'Bi1 RUBBER SHEET IN LARGE STRAIN ANALYSIS' DATABASE CREATE MASTER IDOF=-00111 NSTEP=4 KINEMATICS DISPLACEMENTS=LARGE STRAINS=LARGE ITERATION METHOD=MODIFIED-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=I.E-5 TIMEFUNCTION 1

0.

/

4.

20.9 COORDINATES

/

ENTRIES NODE Y

Z 1 /

2 11.11 TO 5

11.11 1.5 TO 7

9.09 1.5 8

8.08 1.54

/

9 7.07 1.58

/

10 6.06 1.66 11 5.05 1.8

/

12 4.04 2.0

/

1 3 1.03 2.28 14 2.02 2.76

/ 15 1.01 3.42

/

16

0.

4.46 LINE NODES 2 5

/

3 4 LINE NODES 5 16

/

6 TO 15 MATERIAL 1 RUBBER C1=21.605 C2=15.747 EGROUP 1 PLANE STRESS2 GSURFACE 1 2 5 16 EL1=11 EL2=3 NODES=4 EDATA

/

1 0.125 FIXBOUNDARIES 23 INPUT=LINE 1 16 FIXBOUNDARIES 3

INPUT=LINE

/

1 2 LOADS CONCENTRATED 2

2.16666667 3

2.33333333 4

2.33333333 5

2.16666667 SET NSYMBOLS=MYNODES SUBFRAME 12 MESH NNUMBERS=MYNODES ENUMBERS=YES BCODE=ALL MESH CONTOUR=DISTORTION SOLVIA END Version 99.0 Nonlinear Examples Bll1.4

SOLVIA Verification Manual SOLVIA-POST input Bil RUBBER SHEET IN LARGE STRAIN ANALYSIS DATABASE CREATE WRITE FILENAME='bll.lis' SUBFRAME 12 MESH ORIGINAL=DASHED CONTOUR=LPMAX OUTLINE=YES MESH VECTOR=STRETCH EPLINE NAME=A-A 2

4 3

STEP 11 TO 24 4 3 EPLINE NAME=B-B 10 2 1

STEP 11 TO 32 2 1 AXIS ID=i VMIN=-

VMAX=250 LABEL='STRESS-YY' SET PLOTORIENTATION=PORTRAIT FRAME MESH PLINES=ALL SUBFRAME1122 ELINE LINENAME=A-A KIND=SYY SYMBOL=i YAXIS=1 SUBFRAME=2121 ELINE LINENAME=B-B KIND=SYY SYMBOL=i YAXIS=i SUBFRAME=2221 SUBFRAME 12 NHISTORY NODE=2 DIRECTION=2 SYMBOL=i EXYPLOT ELEMENT=-i POINT=i XKIND=EYY YKIND=SYY SYMBOL=-2 ZONE NAME=BB INPUT=ELEMENTS

/

10 21 32 SET ZONENAME=BB NLIST ELIST ELIST SELECT=STRETCH EMAX SELECT=STRETCH GLIST E32 GLIST E32 TSTART=

END Version 99.0 Nonlinear Examples Bl11,5

SOLVIA Verification Manual EXAMPLE B12 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL Objective To verify the PLANE AXISYMMETRIC element in large displacements.

Physical Problem A spherical shell subjected to a concentrated apex load, see figure below, is considered. This problem is also analyzed in Example B65 using the AUTOMATIC-ITERATION method.

R h = 0.01576 in H = 0.0859 in R = 4.76 in z

30 = 10.90 E= 1.0.107 psi v=0.3 Finite Element Model Due to symmetry only one-half of the spherical shell is modelled using ten 8-node PLANE AXISYM METRIC elements as shown in the figure on page B 12.2. The nodes at the apex can only slide in the vertical (Z) direction. At the built-in end the top and bottom nodes can slide in the c-direction of the skew system and the midnode is fixed in b-and c-direction.

Solution Results The input data used in the analysis is shown on pages B 12.4 and B 12.5. The final total apex load is 100 lbs which is reached in 20 equal load steps using Full-Newton iterations.

The deformed mesh for load step 20 is shown in the top figure on page B 12.3. The central deflection as function of the applied apex load for 1 radian is shown in the left bottom figure on page B 12.3. The history of the element stress af, at integration point 7, which is closest to the bottom surface and to the built-in end, is shown in the same figure. The element r-axis is in the 0-direction of the shell.

The apex point load as a function of the deflection ratio Wo, / H is shown in the right bottom figure on page B 12.3.

Version 99.0 Nonlinear Examples BI12.1

,¥

SOLVIA Verification Manual User Hints

The PLANE AXISYMMETRIC element extends 1 radian in the circumferential direction. The total apex load of 100 lbs applied to the actual shell is, therefore, represented by a load of 100/2 applied to the model.

References

[1]

Stricklin, J.A., "Geometrically Nonlinear Static and Dynamic Analysis of Shells of Revolution", High Speed Computing of Elastic Structures, Proceedings of the Symposium of IUTAM, Univ. of Liege, August, 1970.

[2]

Mescall, J.F., "Large Deflections of Spherical Shells Under Concentrated Loads", J. App.

Mech., Vol. 32, pp. 936-938, 1965.

B12 LARCE DISPLACE1EN-ANALYS S OF A SPHERICAL SHELL OR'GINAL H--

O.

7 --1E R

IORG7 0 796 7

3R:*3INAL

-~O.CS NAXES=SKEW rYA S

P I '9.

SO9 I A

SOLV/A-E 990 SOLVIA E.NGTNEERING AB Version 99.0 Nonlinear Examples B 12.2

SOLVIA Verification Manual Nonlinear Examples Bt2 LARGE DISPLACE&ENT ANALYSIS OF A SPHERICAL SHEL ORIGINAL 0

Z 1AX DISPL

[ 0.171z7 T7-ý'E 20 i

.22--

- -T;_

T LOAD 1S.92 SOLVIA-POST 99.0 SOLVTA ENGINEERING AB B12 LARGE DISPLACEMENT ANALYSTS OF A SPHERTCAL SHELL 2

2 16 I1AkJE OF -I'EFUNC-ON L

S4 6

6

!0 I2 64

6 100 S

SLV:A STRICKLIN MESCALL (THIRTY SHELL ELEMENTSI (FINITE DIFFERENCE SOLUTION)

I I 7I LINEAR

/SOLUTION 40 I

I, I-2 20 04 08 12 16 20 DEFLECTION RATIO WO/H IMEFUJC- *0N '.

SOLVTA ENGINEER'NjG AB Version 99.0

,,AL' E 'F OL/IA-POST 99 '

B 12.3

] ] I

,.a

,z H

z

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B12 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL' DATABASE CREATE MASTER IDOF=-00111 NSTEP=20 KINEMATICS DISPLACEMENT=LARGE TOLERANCES TYPE=F RTOL=0.001 RNORM-15.92 ITERATION FULL-NEWTON TIMEFUNCTION 1

0.

/

20.

15.92 SYSTEM 1 CYLINDRICAL COORDINATES ENTRIES NODE R

THETA 1

4.76788

90.

2 4.75212

90.

3 4.75212 79.10 TO 5

4.76788 79.10 SKEWSYSTEM EULERANGLES 1

-10.9 NSKEWS

/

3 1 TO 5 1 MATERIAL 1

ELASTIC E-1.E7 NU-0.3 EGROUP 1

PLANE AXISYMMETRIC GSURFACE 5 1 2 3 EL1=10 EL2=1 NODES-8 SYSTEM=1 FIXBOUNDARIES 2 INPUT=LINES

/

1 2

/

3 5 FIXBOUNDARIES 3 INPUT=NODES

/

4 LOADS CONCENTRATED 1 3 -1.

SET NSYMBOLS=MYNODES HEIGHT=0.25 SMOOTHNESS=YES MESH NNUMBERS=MYNODES VECTOR=LOAD SUBFRAME=12 MESH NAXES-SKEW ENUMBERS=YES BCODE-ALL SOLVIA END Version 99.0 Nonlinear Examples B 12.4

SOLVIA Verification Manual SOLVIA-POST input B12 LARGE DISPLACEMENT ANALYSIS OF A SPHERICAL SHELL DATABASE CREATE STRESSREFERENCE=ELEMENT WRITE FILENAME='bi2.1is' MESH ORIGINAL=DASHED SMOOTHNEES=YES VECTOR=LOAD SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=i DIRECTION=3 XVARIABLE=I SYMBOL=1 OUTPUT=ALL EHISTORY ELEMENT=i POINT=7 KIND=SRR XVARIABLE=i OUTPUT=ALL END Version 99.0 Nonlinear Examples B 12.5

SOLVIA Verification Manual EXAMPLE B13 ELASTIC-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER Objective To verify the PLANE AXISYMMETRIC element for material nonlinear analysis using the thermo elastic-plastic material model subjected to loading, unloading and reloading conditions.

Physical Problem A section of a very long thick-walled cylinder under internal pressure loading at constant temperature, see figure below, is considered. The material is elastic - perfectly plastic.

,Elastic

- perfectly plastic material Plane strain conditions von Mises yield condition E=2.0-10" 1 N/m 2 v=0.3 p

ET =0.0 N/m 2 ca=0.0 1/°C E_

b (y = 400.0.106 N/m 2 at 200C

-, o Gy =352.4.106 N/m 2 at 500C 1y =320.6.106 N/m 2 at 1500C Geometry: a = 0.01 m, b = 0.02 m ay =293.8.106 N/m 2 at 2500C Finite Element Model The model is shown in the figure on page B 13.2. Ten 8-node PLANE AXISYMMETRIC elements are employed for the unit length of cylinder considered. The axial strain is assumed to be zero.

Solution Results The input data is shown on pages B 13.4 and B 13.5. The internal pressure is increased under constant temperature of 20'C to a maximum value of 311.78.106 N/m 2 with an unloading-reloading sequence at 277.14. 106 N/rn2. The stiffness is reformed at every load step. The BFGS method of equilibrium iteration (default) is employed with tolerance ETOL = 10-6.

The radial displacement solution for the nodes at the outer radius and the stress distribution through the wall at the pressure 288.22.106 N/m 2 are shown in the figures on page B 13.3. Excellent agree ment with the solution by Hodge and White [1] is achieved.

Version 99.0 Nonlinear Examples B 13.1

SOLVIA Verification Manual User Hints

" A drastic change in stiffness for the plastic elements occurs at the instant of unloading and solution difficulties can develop if the modified Newton method of iteration is used. However, the BFGS method of iteration, which is the default iteration method, can handle unloading situations success fully and is, therefore, employed.

"* The thermo-elastic-plastic material model can be employed for elasto-plastic analysis (ca=0, Ae=0) as shown here, but the bilinear elastic-plastic material would be more efficient.

Reference

[1]

Hodge, P.E. and White, S.H., "A Quantitative Comparison of Flow and Deformation Theories of Plasticity", Journal of Applied Mechanics, Vol. 17, pp. 180-184, 1950.

B13 EL-ASTIC OFRIINAL 3.002 Ti7NE 0.5 LAS7-C ANALYSIS OF A THICK-vKALLED CY INDER z

y OR-GINA R

PRESSURE 2.309SE7 EOLI,!A-RE 99

ý1"-,

r FAXE-. F SCLV_A ENGINEE.RING A,3 Version 99.0 0.00 z LY R

Nonlinear Examples B 13.2

SOLVIA Verification Manual Nonlinear Examples B !3 E:A,57C PAST>- ANALYSIS 3F

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B13 ELASTIC-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER, DATABASE CREATE MASTER IDOF=-00111 NSTEP=38 DT=0.5 TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

0.

17.

288.22E6 /

COORDINATES ENTRIES NODE 1

2 3

4

6.

277.14E6 /

11.

19.

311.78E6 Y

Z 0.01 0.01 0.01

0.

0.02

0.

0.02 0.01 INITIAL TEMPERATURE TREF=20.

MATERIAL 1

20 2.0E1T 50 2.0E1T 150 2.0E1T 250 2.0E1T THERMO-PLASTIC 0.3 400.0E6 0.3 352.4E6 0.3 320.6E6 0.3 293.8E6 EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1 2 3 4 EL1=1 EL2=10 NODES=8 LOADS ELEMENT INPUT=LINE 1 2 1.

FIXBOUNDARIES 3 INPUT=LINE LOADS TEMPERATURE TREF=20.

SET MESH MESH NSYMBOLS=MYNODES MARGIN=2.0 NNUMBERS=MYNODES ENUMBERS=YES EAXES=RST GSCALE=OLD 23

/

14 VECTOR=LOAD SUBFRAME=21 SOLVIA END Version 99.0 0.

TREF=20

0.

Nonlinear Examples B 13.4

SOLVIA Verification Manual SOLVIA-POST input B13 ELASTIC-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER DATABASE CREATE TOLERANCES WRITE FILENAME='bI3.1is' EPLINE NAME=RADIAL 1 6 5 4 TO 10 6 5 4 SUBFRAME 21 ELINE LINENAME=RADIAL KIND=SXX TIME=17 OUTPUT=ALL SYMBOL=i ELINE LINENAME=RADIAL KIND=SYY TIME=17 OUTPUT=ALL SYMBOL=1 SUBFRAME 21 ELINE LINENAME=RADIAL KIND=SZZ TIME=d7 OUTPUT=ALL NHISTORY NODE=4 DIRECTION=2 XVARIABLE=1 SYMBOL=2 OUTPUT=ALL END Version 99.0 Nonlinear Examples Bi13.5

SOLVIA Verification Manual EXAMPLE B14 THERMO-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER Objective To demonstrate the thermo-elastic-plastic analysis capability of the PLANE AXISYMMETRIC element.

Physical Problem The physical problem is the same as in Example B13 shown on page B 13-1, but the cylinder is now subjected to the pressure and uniform temperature load variation shown in figure below.

p

[N/m2]

300-106 200.106 100.106 277.1 106 242.5 11 0'

242.5 106 r17 3.2 106 202 0 Time 2.

I 6

I I

1 1

1 0 I 2

4 6

8 10 12 14 16 18 20 Temp [Oci 1 00C 20°C 50oc Time 2

4 6

8 10 12 14 16 18 20 Finite Element Model The model is the same as shown on page B 13-2.

Solution Results The input data is shown on pages B 14.3 and B 14.4. The same solution procedure as for the previous Example B 13 is used. The radial displacement response for the nodes on the outer radius and the resi dual stresses at time=10 are shown on page B 14.2.

User Hints 0 The nodal temperatures can be specified directly by the user (as in this example) or they can be read from a file, which may have been generated in a temperature analysis using SOLVIA-TEMP.

Example B38 shows how a combined thermal analysis (using SOLVIA-TEMP) and a thermal stress analysis (using SOLVIA) may be carried out.

Version 99.0 100 80 60 40 20 Nonlinear Examples B 14.1

SOLVIA Verification Manual Nonlinear Examples 31ý THE*V0 -P ASTIC ANALY9--S T.:.

'0 OF A 7H '10K

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B14 THERMO-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER' DATABASE CREATE MASTER IDOF=-00111 NSTEP=40 DT=0.5 ITERATION METHOD=BFGS TOLERANCES ETOL=1.E-3 TIMEFUNCTION 1

0.

/

1. 173.2E6

/

4.

242.5E6

/

8. 242.5E6

9.

240.2E6

/

10. 0.

/

12.

0.

/

13. 240.2E6

20.

277.1E6

/

25.

277.1E6 TIMEFUNCTION 2

0.

20.

/

4.

20.

/

8.

100.

10. 100.

/

12.

50.

/

25.

50.

COORDINATES

/

ENTRIES NODE Y

Z 1 0.01 0.01

/

2 0.01.

/

3 0.02

/

4 0.02 0.01 MATERIAL 1

THERMO-PLASTIC TREF=20 20 2.0E11 0.3 400.0E6

0.

50 2.0E11 0.3 352.4E6

0.

150 2.0E11 0.3 320.6E6

0.

200 2.0E11 0.3 293.8E6

0.

EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1 2 3 4 EL1=1 EL2=10 NODES=8 LOADS ELEMENT INPUT=LINE 1 2

1. 1. 1 FIXBOUNDARIES 3 INPUT=LINE

/

2 3

/

1 4 23/14 INITIAL TEMPERATURES TREF=0.

INPUT=SURFACE 1 2 3 4

20.

LOADS TEMPERATURE INPUT=SURFACE 1 2 3 4

1. 1. 1. 1.

2 SOLVIA END Version 99.0 Nonlinear Examples B 14.3

SOLVIA Verification Manual Nonlinear Examples SOLVIA-POST input B14 THERMO-PLASTIC ANALYSIS OF A THICK-WALLED CYLINDER DATABASE CREATE WRITE FILENAME='b14.1is' EPLINE NAME:RADIAL 1

6 5 4 TO 10 6 5 4 SUBFRAME 21 ELINE RADIAL KIND=SXX TIME=-0 OUTPUT=ALL ELINE RADIAL KIND=SYY TIME=-0 OUTPUT=ALL SUBFRAME 21 ELINE RADIAL KIND=SZZ TIME=10 OUTPUT=ALL NHISTORY NODE=4 DIRECTION=2 XVARIABLE=1 SYMBOL=2 OUTPUT=ALL ELIST SUMMATION KIND=REACTION END Version 99.0 B 14.4

SOLVIA Verification Manual EXAMPLE B15 CYCLIC CREEP ANALYSIS OF A THICK-WALLED CYLINDER Objective To verify the creep material model when employed with PLANE AXISYMMETRIC elements.

Physical Problem The figure below shows the thick-walled cylinder which is to be analyzed for cyclic internal pressure.

The pressure is varied as shown in the figure. Plane strain conditions are assumed. The material of the cylinder is assumed to obey a uniaxial creep law of the form F, = 6.4. 10-L "* (74.4.t Plane strain conditions E = 2.0-107 lb/in2 v=0.3 Temperature 100°F Geometry a = 0.16 in b = 0.25 in 2.2 2.6 t(Hr)

-36 5 Finite Element Model The model is shown in the top figure on page B 15.3. It consists of eight PLANE AXISYMMETRIC 8-node elements.

Version 99.0 365 P

(lb/In2_

B 15.1 9.

h L

Nonlinear Examples

-Y

SOLVIA Verification Manual Solution Results The input data is shown on pages B 15.4 and B 15.5. The BFGS method of equilibrium iteration is used with tolerance ETOL = 0.000 1 and the stiffness matrix is reformed at every step. The creep rate equation is integrated using 10 subdivisions and the Euler forward method.

The left bottom figure on page B 15.3 shows the effective stress at the element integration points closest to the inside and outside surfaces of the cylinder. The solution results compare well with the steady state solution reported by Bailey [1].

The right bottom figure on page B 15.3 shows the von Mises stress distribution in the cylinder in the last solution step and the radial displacement history of the inner cylinder radius.

User Hints

" The explicit method of integrating the creep equation requires a subdivision of the time step into sufficiently small increments for stability reasons. The implicit integration schemes can display considerably better stability characteristics, but then also iterations are used at the integration points for the creep strains.

The time step At is here selected using the "rule of thumb" that the maximum creep strain increment must be smaller than 1/ a=

1 a

1/ 4.4 of the current elastic strain.

Reference

[1]

Finnie, I., and Heller W.R., Creep of Engineering Materials, McGraw-Hill, 1959, p. 208.

Version 99.0 Nonlinear Examples B 15.2

SOLVIA Verification Manual Nonlinear Examples BI5 CYCLIC CREEP ANALYS S OF A TRICK-UALLED CYLINDER z LY R

ORIGINAL I-0.2 PRESSURE 36-5 SOLVIA-PRE 99.0 SOLVIA ENGINEE

>4

-

BIS CYCLIC CREEP ANALYSIS OF A THICK-WALLED C YLINDER MAX DSPL. --

9.296!E-6 Z

TIME 5.8 Y

MISES MAX 789 53 S779 20 758 54

,37 88 7t7 22 S696. 57 675 S[

6SN5 2S S634.59 MIN 621426

1.

T OI 71, <RAJRS SOL'I:A ENGINEERING AB SOLVIA-PDST 99 0 SOLV A ENGINEERING AB Version 99.0 ORIG NAL TIME 0, I

0. 02 B15 CYCLIC CREEP ANALYSIS OF A THICK-WALLED CYLINDER MASTER 100'11:

B 10i1:1 RING AB A -A-I-SCLVfA-OST 99.4 i

i uJ w

u_

B 15.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B15 CYCLIC CREEP ANALYSIS OF A THICK-WALLED CYLINDER' DATABASE CREATE MASTER IDOF=I00111 NSTEP=48 DT=0.1 TOLERANCES ETOL=1.E-4 ITERATION METHOD=BFGS TIMEFUNCTION 1

0.

365.

/

2.2 365.

2.6 -365.

/

10.

-365.

TIMEFUNCTION 2

0. 1. /

100. 1.

COORDINATES

/

ENTRIES 1

.16

.1 /

2

.16 NODE Y

Z 3

.25

/

4

.25

.1 INITIAL TEMPERATURES TREF=00-.

MATERIAL

1. PLASTIC-CREEP XKCRP=1 TREF=i00.

ALPHA=C.,

AC=6.4E-18 A1=4.4 A2=1.

70.

2.E7 0.3 8.E7 2.E6 0.

100.

2.E7 0.3 8.E7 2.E6 0.

EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1 2 3 4 EL1=1 EL2=8 NODES=8 LOADS ELEMENT INPUT=LINE 1 2

1. 1. 1 FIXBOUNDARIES 3 INPUT-LINE

/

2 3

/

1 4 LOADS TEMPERATURES INPUT=SURFACE 1 2 3 4 100.

100.

2 MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES MESH ENUMBERS=YES BCODE=ALL SOLVIA END VECTOR=LOAD SUBFRAME=21 Version 99.0 Nonlinear Examples B 15.4

SOLVIA Verification Manual SOLVIA-POST input B.15 CYCLIC CREEP ANALYSIS OF A THICK-WALLED CYLINDER DATABASE CREATE WRITE FILENAME='blS.lis' AXIS ID=1 VMIN=-.

VMAX=4.8 LABEL='TIME IN HOURS' AXIS ID=2 VMIN=O.

VMAX=1200.

LABEL='EFFECTIVE STRESS' SET PLOTORIENTATION=PORTRAIT EHISTORY ELEMENT=i POINT=9 KIND=MISES XAXIS-1 YAXIS-2, SYMBOL=2 SSKIP=i EHISTORY ELEMENT=8 POINT=i KIND=MISES XAXIS=-i YAXIS=-2, SUBFRAME=OLD OUTPUT=ALL MESH CONTOUR=MISES SUBFRAME=12 NHISTORY NODE=i DIRECTION=2 OUTPUT=ALL SYMBOL=i SSKIP=2 END Version 99.0 Nonlinear Examples B 15.5

SOLVIA Verification Manual EXAMPLE B16 LARGE DISPLACEMENT ANALYSIS OF A PLATE Objective To verify the SHELL element for large displacement analysis.

Physical Problem A simply supported square plate subjected to uniform pressure load, see figure below, is considered.

The plate material is linear elastic.

ONE 16-NODE E = Young modulus = 107 psi ELEMENT'" "

v = Poisson ratio = 0.1 h = Plate thickness = 0.12 in a = Plate width = 24 in q = Uniform applied pressure psi 4 x 4 x 2 Gauss integration Finite Element Model Because of symmetry conditions only 1/4 of the plate need be modeled, see top figure on page B 16.3.

One 16-node SHELL element is used. Constraint equations are used to model uniform in-plane edge displacements.

Solution Results The input data is shown on page B 16.4. The loading is applied in 22 load steps up to the final pressure q = 3.125 psi. For every load step the stiffness is reformed and the BFGS method of equilibrium iteration is employed with tolerance ETOL = 1.10-8.

The computed center deflection ratio w/h as a function of the load parameter K is shown in the figure on page B 16.2. The computed displacement response agrees very closely with the solution given by Levy [1]. The deformed shape corresponding to the final load step and center deflection as a function of the pressure are shown in the left bottom figure on page B 16.3.

A contour plot of the vertical displacement and a vector plot of the principal stresses are shown in the right bottom figure on page B 16.3.

User Hints The large displacement response of shells generally involves severe nonlinearities because the structure, initially only in bending action, is subjected to increasing membrane action with increasing load. This results into a stiffening effect. The BFGS method is generally quite effective for these types of problems. If difficulties are encountered in the convergence of iteration, then the powerful full Newton iteration with line searches can be employed.

Version 99.0 Nonlinear Examples B 16.1

SOLVIA Verification Manual Reference

[1]

Levy, S., "Bending of Rectangular Plates with Large Deflections", Technical Notes, National Advisory Committee for Aeronautics, No. 846, 1942.

5.0 o LINEAR SOLUTION 0

z Uj I U

LU 0

I Cr rr z

LU U

4.0 3.0 2.0 1.0 UNIFORM IN-PLANE EDGE DISPLACEMENT PRESENT STUDY LEVY 0

100 LOAD PARAMETER K = qa 4 /Eh 4 Version 99.0 Nonlinear Examples B 16.2

SOLVIA Verification Manual Nonlinear Examples 306 LARGE DISPLACEMENT ANALYSTS OF A PLATE x

PRESSURE

3. 125 0 0 0

C 013 1 D C02C00 F

Ot33II F : 1011 1 G !2101 1 H 2010CC I 21 11C SCLV/IA ENGINEERING AB SCLVIA-PRE 99.0 816 LARGE DISPLACEMENT ANALYSFS OF A 'LATE R'GINAL

,1 2.

MAX DISPL.

C 4388 TIME I

z C

8REACT ION 8554.5

'C--

VAjL') F T IMFFNCT N

SOLVIA-POST 99 0 SOLVIA ENGINEERING AB B[6 LARGE DISPLACEMENT ANALYSIS OF A PLAT:

MAX DISPL. H 0.4388 TINE i z

x '-y Z-D[R CISPL 0AX C.43880

'0.4i 38 0 38653 C 30168 0 2 683 ii0.9198 1 193 S0. 1 3713 3 0382227S 30.027042 MIN 0 X

F MAX DCSPL.

C '388

-.!ME I SPRINCCPAL SHELL TOP SOLVIA-pOST 99.0 9496.

9496 2 SCLVIA ENGINEERUNG 3B Version 99.0 I

-Z B 16.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'Bi6 LARGE DISPLACEMENT ANALYSIS OF A PLATE' DATABASE CREATE MASTER NSTEP-20 DT=0.05 KINEMATICS DISPLACEMENTS=LARGE TOLERANCES ETOL=1.E-8 ITERATION METHOD=BFGS TIMEFUNCTION 1

0.o0.

1.

3.125 COORDINATES 1

0.

12.

TO 4

To 7

12.

/

8

12.

MATERIAL 1

ELASTIC E=1.E7 NU=0.316228 EGROUP 1 SHELL GSURFACE 1 4 7 8 THICKNESS 1

0.12 LOADS ELEMENT 1 -T 1.

FIXBOUNDARI ES FIXBOUNDARI ES FIXBOUNDARI ES CONSTRAINTS 1 1 4 1 1.

5 2 4 2 1.

EL1=1 EL2=1 NODES=-6 36 INPUT=LINE 156 INPUT=LINE 246 INPUT=LINE TO 3 1 4 1 1.

TO 7 2 4 2 1.

MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL TIME=1.0 SOLVIA END VECTOR=LOAD, SOLVIA-POST input B16 LARGE DISPLACEMENT ANALYSIS OF A PLATE DATABASE CREATE WRITE FILENAME='bI6.1is' SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 MESH ORIGINAL=DASHED NSYMBOLS=MYNODES VECTOR=REACTION NHISTORY NODE=8 DIRECTION=3 XVARIABLE=1 SYMBDL=1 OUTPUT=ALL SUBFRAME 12 MESH CONTOUR:DZ MESH VECTOR=SPRINCIPAL SUMMATION SUMMATION END KIND=REACTION KIND=LOAD Version 99.0

/ /

/

1 4 7 8 8 1

/

4 7 Nonlinear Examples B 16.4

SOLVIA Verification Manual EXAMPLE B17 THERMO-ELASTIC ANALYSIS OF A CANTILEVER BEAM Objective To verify the thermo-elastic material model when used with SOLID elements.

Physical Problem A cantilever beam is subjected to a linearly varying temperature distribution in the Z-direction, see figure below. No other loads are applied to the beam. The same physical problem is also analyzed in Example B38, but using temperatures calculated by SOLVIA-TEMP.

Z a

150' NEUTRAL AXIS

__21a L

70OF 200°F L = 6 in.

E = 3.0.107 lb/in 2 E = 3.0.10 7 lb/in 2 a= 1 in.

v=0.3 v=0.3 ca = 6.0.10-6 in/in/°F c = 6.0.10- 6 in/in/°F Reference temperature = 125°F Finite Element Model Since the problem is anti-symmetric with respect to the neutral axis of the beam, only the portion above the neutral axis is modeled, see figure on page B 17.2. Three 20-node SOLID elements are used.

Solution Results The input data is shown on pages B 17.3 and B 17.4. The analysis is carried out for material nonlin earities only. The analytical solution for the vertical displacement component using the theory given, for example, in [1], is W = _y2.1.5 4 (in)

The vertical displacement component at the free end of the beam, Y = 6 (in), is:

Theory SOLVIA (node 1)

-5.400. 10

-5.400.10-'

An excellent agreement can be observed. The deformed mesh and contour lines of temperature are shown in the figure on page B 17.3. Note that the maximum displacement given in the figure is the maximum total displacement. not the maximum component displacement.

Version 99.0 Nonlinear Examples B 17.1

SOLVIA Verification Manual User Hints The thermo-elastic material model allows Young's modulus E, Poisson's ratio v and the mean co efficient of thermal expansion ct to vary as a function of temperature. The material model is con sidered nonlinear since the resulting stiffness matrix, in general, is dependent on the applied tem perature loading. However, no equilibrium iterations are necessary except in analysis with one (or both) of the two conditions:

1)

The stiffness matrix is not reformed in every step.

2)

The model accounts for large displacement effects.

The reference temperature TREF is the temperature for which the thermal stresses are zero. Note that TREFN serves as the reference temperature for the mean value of the coefficient of thermal expansion ca.

Since the theoretical displacement solution varies quadratically in the X-and Y-directions, the 20 node SOLID element can describe the displacement field exactly. Therefore, since the beam is free to bend in both the X-and Y.directions, the stresses are zero everywhere.

The problem can also be modeled using BEAM or ISOBEAM elements since these elements accept thermal gradient loadings in the transverse directions.

Reference

[1]

Boley, B.A., and Weiner, J.H., Theory of Thermal Stresses, John Wiley and Sons, 1960, pp. 307-314.

B17 7HERMO-ELAST7C ANALYS7S O

A CANTILEVER BEAM ORCGINAL O.S Z

X B

G 1' 1

3t_0 '_ 1 SOLA A. :3IN AB SOLV!A ENGINEERING AB CoL'/7A-P E 99.3 Version 99.0 Nonlinear Examples B 17.2

SOLVIA Verification Manual B17 THERC-ELASTIC ANALSIS rF A CANT7LEVER BEAN 11AX DISPL 5-.47-SE-3 T711 1

Z x

y-TEMPERATURE MAX 1SO.00 1 2.

88 135,938

32. 813 129.688 1'

26.S63 MIN 12S.000 SOLVIA ENGINEERING AB SOLVIA-POST 99.0 SOLVIA-PRE input HEADING

'B17 THERMO-ELASTIC ANALYSIS OF A CANTILEVER BEAM' DATABASE CREATE MASTER IDOF=000111 COORDINATES 1

0.

6.

0.5

/

2

0.

0.5 3

1. 0.

0.5

/

4

1. 6.

0.5 5

0.

6.

0.

/

6

0.

7

1. 0.

0.

/

8

1. 6.

0.

MATERIAL 1

THERMO-ELASTIC TREF=125.

70 3.E7 0.3 6.E-6 200 3.E7 0.3 6.E-6 EGROUP 1

SOLID GVOLUME 1

2 3 4 5 6 7 8

ELI=3 EL2=1 EL3=1 NODES=20 FIXBOUNDARIES 12 INPUT=SURFACE

/

2 3 7 6 FIXBOUNDARIES 3

INPUT=LINE

/

6 7 FREEBOUNDARIES 13 INPUT:LINE

/

2 6

/

3 7 FIXBOUNDARIES 12 INPUT=SURFACE

/

5 6 7 8 LOADS TEMPERATURE TREFN=125 INPUT=ZONE INTERPOLATION=Z, COORD1=0.

V1=0.

COORD2=0.5 V2=25 ZONEI=WHOLE MESH NSYMBOLS-MYNODES NNUMBERS=MYNODES BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B 17.3

SOLVIA Verification Manual SOLVIA-POST input B17 THERMO-ELASTIC ANALYSIS OF A CANTILEVER BEAN DATABASE CREATE WRITE FILENAME='b17.1is' MESH CONTOUR=TEMPERATURE ORIGINAL=YES NLIST ZONENAME=EL1 END Version 99.0 Nonlinear Examples B 17.4

SOLVIA Verification Manual EXAMPLE B18 ANALYSIS OF AN UNDERGROUND OPENING Objective To demonstrate the use of the curve description material model in a simplified analysis of an underground opening.

Physical Problem A tunnel through a rock mass, see figure below, is considered. The walls of the tunnel are reinforced with rock bolts. The rock material is assumed to have constant Young's modulus and Poisson's ratio.

Rock material (E and v constant) 0 E

43200 kip/ft2 v=0.

P SCurve description material model L l 1 L

1

_ 60 Material weight density for tensile failure option 0.12 kip/ft3 76 Normal stiffness reduction factor 1.10-3 T

80 Shear stiffness reduction factor 9.9.10-4 86 Bulk modulus 14400 kip/ft2 Loading shear modulus 21600 kip/ft2 9t, Rock bolts 104 E = 4.32.106 kip/ft2 Yield= 14400 kip/ft2

_116

)/////*/////'//

////////ET

=0.

Area = 0.0021 ft2 32 136P1 40 60 Y

Initial strain 1.103 Finite Element Model The model is shown in the left top figure on page B 18.3. The rock is modeled using 8-node PLANE STRAIN elements and the rock bolts by 2-node TRUSS elements. The pressure due to the overburden rock material is simulated by element pressure loads. The curve description model is employed to describe the rock material. The tensile failure option is used with material density y = 0.12 kip/ft3,

and shear reduction factor = 0.00099. Both the tensile stress and the shear stress are, therefore, reduced to zero for a crack, i.e. when any principal tensile stress exceeds the in-situ gravity stress at an integration point.

Version 99.0 Nonlinear Examples B 18.1

SOLVIA Verification Manual Solution Results The input data is shown on pages B 18.4 and B 18.5. A materially-nonlinear-only analysis is carried out in eleven steps using the full Newton iteration with line search. The final overburden pressure reached in the analysis is 30 kip/ft 2 and the corresponding distribution of Tresca effective stress and of the hydrostatic pressure (negative mean stress) are shown in the bottom figure on page B 18.3.

The displacement histories at two characteristic locations are shown in the right top figure on page B 18.3.

User Hints

" When cracks form in a finite element model using the curve description material model (or the concrete model) the stiffness generally changes drastically. Depending on the in-situ gravity pressure the change in stresses associated with the formation of a crack may also be significant.

These effects generally result into a considerable redistribution of stresses due to cracking.

It is, therefore, important to take small enough load steps whenever cracks are forming in order to avoid solution difficulties.

"* Note that the in-situ gravity pressure p, at the nodes is calculated using Pi =y.Zi where Z1 is the nodal Z-coordinate. The stress corresponding to this gravity pressure is not included in the calculated stresses which are output from SOLVIA. It is, however, taken into account when the bulk and shear moduli are determined and in the criteria for cracking.

Version 99.0 Nonlinear Examples B 18.2

SOLVIA Verification Manual Nonlinear Examples S8' ANALYSIS OF AN UNDERGROUND OPENING 5

i 0 s

2 2

3G VALUE OF TIMEF-NCTION 0.

1s I5 20 25 30 VWLUE OF TDrEFUNCT:ON t

SOLVIA-POST 99.0 SOLVIA ENGINEERING AB B18 ANALYSIS ORIGINAL

10.

MAX DISPL.

0.096793 T!ME it OF AN UNDERGROUND OPENING Z

ORIGINAL

10.

MAX DiSPL. ý--

0.096793 L TIME 11 TRESOA MAX !:' 6.54 S94.68S S80. 1 18 S...'

65.55 1.

so S

984

36. 41 7 S2'.. 8S0

7. 835 MIN SOLVIA-POST 99 0

/*/

MEAN S-RESS MAX 2.207

-6.0367

-ii

-535

-22 532

-28. 030

-33.28

,-1 39,027 MIN-4 A.776 SOLV7-A ENGINEERING AB Version 99.0 y

LY I'

I B 18.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B18 ANALYSIS OF AN UNDERGROUND OPENING' DATABASE CREATE MASTER IDOF=100111 NSTEP=11 ITERATION FULL-NEWTON LINE-SEARCH=YES TIMEFUNCTION 1

0.

/

1. 20.

/

11. 30.

COORDINATES

/

ENTRIES NODE Y

Z 1

60.

-116.

/

2

60.

-104.

/

3

32.

-60.

TO 7

32.

-76.

TO 9

32.

-86.

TO 11

32.

-94.

TO 13

32.

-104.

TO 15

32.

-116.

16

48.

-60.

TO 20

48.

-76.

TO 22

48.

-86.

/

23 43.75 -89.75 24

40.

-94.

/

25 37.75 -98.75 /

26

36.

-104.

TO 28

36.

-116.

29

60.

-60.

TO 33

60.

-76.

TO 35

60.

-80.

/

36 53.75 -82.5 NGENERATION NSTEP=34 YSTEP=-32.

/

3 TO 15 LINE NODES 3

15

/

4 TO 14 LINE NODES 16 28

/

17 TO 27 LINE NODES 16 22

/

17 TO 21 LINE NODES 29 35

/

30 TO 34 LINE NODES 37 49

/

38 TO 48 LINE NODES 35 22

/

36 MATERIAL 1

PLASTIC E=4.32E6 YIELD=1.44E4 ET=0.

MATERIAL 2

CURVE-DESCRIPTION ICRACK=2 GAMMA=0.12 STIFAC=1.E-3, SHEFAC=9.9E-4

.0 1.44E4 1.44E4 2.16E4

/

.01 1.44E4 1.44E4 2.16E4

.05 1.44E4 1.44E4 2.16E4

/

.1 1.44E4 1.44E4 2.16E4

.5 1.44E4 1.44E4 2.16E4

/

1.

1.44E4 1.44E4 2.16E4 EGROUP 2

PLANE STRAIN MATERIAL=2 GSURFACE 37 49 15 3 EL1=6 EL2=2 NODES=8 GSURFACE 3

15 28 16 ELI=6 EL2=1 NODES=8 GSURFACE

.16 22 35 29 EL1=3 EL2=1 NODES=8 GSURFACE 26 28 1 2

EL1=1 EL2=2 NODES=8 LOADS ELEMENT INPUT=LINES 37 29 1.0 EGROUP 1

TRUSS MATERIAL=1 ENODES

/

1 22 9

/

2 22 20

/

3 35 33 EDATA

/

1 2.1E-3 1.E-3 TO 3 2.1E-3 1.E-3 FIXBOUNDARIES 23 INPUT=NODES

/

47 TO 49 FIXBOUNDARIES 3

INPUT=LINES

/

49 15

/

15 28

/

28 1 FIXBOUNDARIES 2

INPUT=LINES

/

1 2

/

35 29 SET PLOTORIENTATION=PORTRAIT NSYMBOLS=MYNODES VIEW=X MESH NNUMBERS=MYNODES VECTOR=LOAD SUBFRAME=12 MESH ENUMBERS-EGI BCODE=ALL OUTLINE=YES NSYMBOL=EG1 SOLVIA END Version 99.0 Nonlinear Examples B 18.4

SOLVIA Verification Manual SOLVIA-POST input B18 ANALYSIS OF AN UNDERGROUND OPENING DATABASE CREATE WRITE FILENAME='bI8.1is' SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=35 DIRECTION=3 XVARIABLE=1 SYMBOL=i OUTPUT=ALL NHISTORY NODE=26 DIRECTION=3 XVARIABLE=I SYMBOL=2 OUTPUT=ALL SET VIEW=X ORIGINAL=DASHED OUTLINE=YES SET PLOTORIENTATION=LANDSCAPE MESH CONTOUR=TRESCA SUBFRAME-21 MESH CONTOUR=SMEAN END Version 99.0 Nonlinear Examples B 18.5

SOLVIA Verification Manual EXAMPLE B19 REINFORCED CONCRETE BEAM, PLANE STRESS Objective To demonstrate the concrete material model in plane stress analysis and the modeling of reinforced concrete.

Physical Problem A simply supported reinforced concrete beam subjected to two symmetrically applied concentrated loads as shown in the figure below is considered. The steel reinforcement is located 2.06 in. from the bottom surface of the beam.

Pi2 P/2 PLANE OF CONTRAFLEXURE PLANE OF CONTRAFLEXURE BEAM DIMENSIONS Material parameters CONCRETE:

Density Initial tangent modulus Poisson's ratio Uniaxial cut-off tensile strength Uniaxial maximum compressive stress (SIGMAC)

Compressive strain at SIGMAC Uniaxial ultimate compressive stress Uniaxial ultimate compressive strain STEEL:

Density Young's modulus Initial yield stress Strain hardening modulus 0.2172.103 3060 0.2 0.458

-3.74

-0.002

-3.225

-0.003 0.7339. 10 30000 44 300 lbf s2/in4 ksi ksi ksi in/in ksi in/in lbf. s 2/in4 ksi ksi ksi Finite Element Model The finite element model used is shown in the figures on page B 19.4. Using symmetry conditions only one-half of the structure need to be considered. Forty 8-node PLANE STRESS elements are used to model the concrete of the beam and ten 3-node TRUSS elements are used to model the reinforce-Version 99.0 Nonlinear Examples B 19.1

SOLVIA Verification Manual ment. In addition, three vertical TRUSS elements are used at the left end support of the beam to distribute the support force. The material properties of the concrete are idealized using the concrete material model. The material of the steel bars is modeled as elastic-plastic with strain hardening. The solution is obtained by using the AUTOSTEP method and full Newton iteration with line searches.

Solution Results The input data on pages B 19.9 and B 19.10 is used in the finite element analysis.

Krahl et al. [1] analyzed the development of cracks in reinforced concrete beams subjected to pure moment. Using a simplified stress-strain law for the concrete and excluding the tension stiffening effect after tensile cracking the critical (collapse) load was found to be approximately 10 kips for the beam in this example with reinforcement area 0.62 in2. Suidan and Schnobrich [2] presented a finite element solution of the same beam using the simplified stress-strain law for the concrete.

The SOLVIA solution results can be seen on pages B 19.5 and B 19.6. The collapse load of the beam is calculated by the model to be 10.5 kips. The SOLVIA solution compares well with the results found in [1] and [2].

The top figure on page B 19.5 shows the formation of cracks as contour and vector plots for the last load step of the analysis. The bottom figures on page B 19.5 show the midspan deflection of the beam as a function of the load together with typical stress-strain curves for the concrete material in tension and compression and the elastic-plastic response curve from the steel reinforcement.

The figures on page B 19.6 show the stress and strain variations along typical lines in the concrete and along the reinforcement bar. It can be seen that the tensile stresses along the line SECTION are small due to cracking. The distribution of axial strain along the line BAR in the reinforcement and along the line ABOVE in the concrete shows a concentration of strain in the line portion located under the load application.

An analysis was also carried out for the beam when the reinforcement area is A, = 2.0 in 2 and corresponding results are shown on pages B 19.7 and B 19.8. The analysis is stopped because of crushing of the concrete under the load application. Since the reinforcement remains elastic strain concentration due to cracking is avoided.

User Hints

"* The solution of concrete problems can be difficult, due to the sudden nonlinearities that take place as a result of cracking and crushing of the material. The overall structural nonlinearities are more pronounced when only small amounts of steel reinforcements are used in the structure.

" When the structure is lightly reinforced the effect of the nonlinearities can be sudden and signifi cant strain redistributions can occur for each load increment. Note that cracking is associated with a negative slope in the so-called tension stiffening portion of the uniaxial stress and strain relation for concrete. If enough cracking takes place in a structure the total tangent stiffness matrix may also be negative so that local maxima (and minima) may occur in the force-displacement response.

Conventional iteration methods can then not be used and the AUTOMATIC-ITERATION method need then be tried, unless a remodelling is performed.

Version 99.0 Nonlinear Examples B 19.2

SOLVIA Verification Manual

" The redistribution of strain during cracking can be mesh and path dependent. While some integra tion points may show strain concentration, thus in simplified words move along the uniaxial stress and strain curve with decreasing stress for increasing strain (the tension stiffening portion), other integration points may unload with decreasing stress and strain. This possibility of different stress unloading paths when cracking has occurred can cause numerical difficulties.

" In reinforced concrete analysis it may in many cases be important to make a relatively coarse mesh so that the overall structural behaviour is described rather than all the details. The amount of details in the complex world of reinforced concrete analysis can be overwhelming and can, if included in the model, very well make the solution difficulties insurmountable in practice.

It is often of interest to verify the concrete material parameters by running uniaxial tension and compression tests using a single element and compare the produced stress-strain curves with experimental curves, if available. It is particularly important to verify the effect of the chosen values for 7 / sc and oy / E. on the uniaxial compression curve.

" Convergence of the equilibrium iterations is often improved by specifying as large values for KAPPA, SHEFAC and the equilibrium tolerances as are technically reasonable.

" In case of severe difficulties in obtaining a converged solution one may often gain insight in the behaviour of the model by running an analysis with very small load steps and without equilibrium iterations, see command EQUILIBRIUM-ITERATION in the SOLVIA-PRE Users Manual.

References

[1]

Krahl, N.W., Khachaturian, M., and Seiss, C.P., "Stability of Tensile Cracks in Concrete Beams", ASCE, J. Struct. Div., Feb. 1967, pp. 235-254.

[2]

Suidan, M., and Schnobrich, W., "Finite Element Analysis of Reinforced Concrete", ASCE, J. Struct. Div., Vol. 99, No. ST1O. Oct. 1973.

Version 99.0 Nonlinear Examples B 19.3

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SOLVIA Verification Manual EXAMPLE B20 CONCRETE MATERIAL CURVES IN COMPRESSION Objective To verify the uniaxial and biaxial compression characteristics of the SOLVIA concrete material model.

Physical Problem In a classical paper by Kupfer et al. [1], concrete specimens of dimension 20 x 20 x 5 cm were sub jected to stress combinations in the regions of biaxial compression, compression-tension and biaxial tension. In this example the SOLVIA concrete material model is compared with the test results of Kupfer et al. in the biaxial compression and compression-tension regions including uniaxial compression.

Finite Element Model One 4 node PLANE STRESS2 element is used for the uniaxial case and one 8 node SOLID element is used in the other analyses. The biaxial loading is applied as element pressure in the global Y-and Z directions. The X-direction is stress-free.

Kupfer et al. analyzed three types of concrete with an unconfined uniaxial compressive strength of 190, 315 and 590 kp/cm 2.

The material data used in this example is based on the experimental results for the concrete with the compressive strength 315 kp/cm 2. All material parameters are shown on the pages with input data and include:

a,

-315 kp/cm 2 g0 =-0.0021

ý= -280 kp/cm 2 Eu= -0.0031 E= 325000 kpicm 2 v =0.2 d, = 28.35 kp/cm2 uniaxial compressive strength adjusted uniaxial strain value corresponding to the compressive strength ultimate compressive stress corresponding ultimate strain Young's modulus Poisson's ratio uniaxial tensile strength (- 0.09 -

)

The parameters of the failure surface in compression when cy,

= 0 are:

SP31I=I.0 yzz /

when (5v =-0 SP312=t.24 (7, / a, when a7yy = 0.5(cizz SP313=1.15 o, /a, when Cyy = Uzz The experimental failure envelope and the failure envelope used in SOLVIA are shown below together with the four stress load paths used in the calculations.

Version 99.0 Nonlinear Examples B20.1I

SOLVIA Verification Manual Kupfer et al [1]

SOLVIA

Tensile failure envelope Compression failure envelope The Automatic-Iteration method is used to trace the pre-and post-crushing behaviour of the concrete model. Force tolerance is used in the equilibrium iterations with RNORM = 1 and RTOL = 0.001.

The finite element model is shown in the top left figure on pages B20.4 to B20.7 for the analysis cases.

Solution Results The first analysis is the uniaxial case with loading only in the Z-direction and the input data used is shown on pages B20.8 and B20.9. A plane stress finite element model, see top left figure on page B20.4, was employed but the 3D model with a SOLID element would yield the same results. The stress response in the load direction as a function of the strain in the same direction and the strain in the transverse direction are shown in the top right figure on page B20.4. The symbols show the corre sponding experimental results from [1].

The stress in the load direction is also shown as a function of the volumetric strain e, + eyy + e=

(left bottom figure). The agreement with the experimental results of [11 shown as symbols is good up to 80-90 % of the concrete compressive strength. At this load level microcracks start to form in the experiment which is accompanied with a significant increase of the volumetric strain.

The strain in the load direction as a function of the volumetric strain is shown in the right bottom figure of page B20.4 together with the corresponding experimental values. A constant Poisson's value gives a linear relationship as obtained in the analysis. In the experiment the relationship is linear up to the load level 80-90% of the compressive strength where microcracks start to develop.

In the analysis case B20A equal biaxial compressive loading is used, see the input data on pages B20. 10 and B20. 11 and the figures on page B20.5. The top right figure shows a7, as a function of the strain e, and e The X-direction has no applied pressure. The strain response in the Y-direction is equal to the strain response in the Z-direction, as it should, see the left bottom figure. The volumetric strain response is shown in the right bottom figure. In all three figures the corresponding experimental Version 99.0 Nonlinear Examples B20.2

SOLVIA Verification Manual results are shown as symbols and the response earlier calculated in the uniaxial case is also shown for comparison. Good agreement with the experimental values can be seen up to 80-90% of the com pressive strength. Higher load levels result again in a significant increase in the experimental volu metric strain which cannot be reproduced since Poisson's ratio is constant in the analysis.

In the analysis case B20B the compressive loading in the Y-direction is 0.52 times the loading in the Z-direction, see the input data on pages B20.12 and B20.13 and the response figures on page B20.6.

The strain responses e,, and e,, are shown in the top right figure and e and e in the left bottom figure together with the corresponding experimental results plotted as symbols. The volumetric strain response is shown in the right bottom figure. The corresponding uniaxial strain responses as earlier calculated are shown in the three figures for comparison. Good agreement between the calculated and the experimental values can again be seen up to the load level of 80-90% of the compressive strength.

In the analysis case B20C the Y-direction load is -0.103 times the compressive loading in the Z-direc tion. The Y-direction load is then in tension. The input data is shown on pages B20.14 and B20.15 and the response figures on page B20.7. The experimental results are marked as symbols in the fig ures and the uniaxial curves have been included for comparison. The calculations stopped when the failure surface was reached at cyy = 17.1 kp / cm 2 and

, = -166.3 kp / cm 2. Failure was reached in the experiment when

, = -194 kp / cm2.

Reference

[1]

Kupfer, H., Hilsdorf, H.K. and Rusch, H., "Behaviour of Concrete Under Biaxial Stress", Title No. 66-52, ACI Journal, Aug. 1969.

Version 99.0 Nonlinear Examples B20.3

SOLVIA Verification Manual Nonlinear Examples 820 CONCRETE MATERIAL CURVES IN COMPRESSON.

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SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING

'B20B CONCRETE MATERIAL CURVES IN COMPRESSION, 0.52*SZZ=SYY' DATABASE CREATE MASTER IDOF=000111 NSTEP=I00 DT=I.

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MATERIAL 1

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STIFAC=0.0001 SHEFAC=0.5, SP311=I.0 SP312=1.24 SP313=1.15 EGROUP 1

SOLID ENODES 1

12345678 LOADS ELEMENT 1

R 0.52 0.52 1

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/

1 2 5 6 FIXBOUNDARIES 2

/

2 3 6 7 FIXBOUNDARIES 3

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'B20C CONCRETE MATERIAL CURVES, COMPRESSION-TENSION,

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0 10 20

/

2 0 0 20

/

3 5 0 20

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4 5 10 20 5

0 10 0

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6 0

0 0

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7 5 0 0

/

8 5 10 0

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CONCRETE EO=325000 NU=0.2 SIGMAT=28.35, SIGMAC=-315 EPSC=-0.0021 SIGMAU=-280 EPSU=-0.0031, BETA=0.5 KAPPA=15.

STIFAC:0.0001 SHEFAC=0.5, SP311=1.0 SP312=1.24 SP313=1.15 EGROUP 1

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SOLVIA Verification Manual EXAMPLE B21 LARGE DEFLECTION ANALYSIS OF A SHALLOW ARCH Objective To verify the large displacement behaviour of the ISOBEAM beam element and demonstrate the restart option in static analysis.

Physical Problem A shallow arch subjected to a concentrated apex load, as shown in the figure below, is considered.

R = 133.144 in h = 0.1875 in b = 1.00 in (width)

= 7.33970 E = 107 psi Rtw v=0.2 Finite Element Model The finite element model is shown in the figure on page B21.2. Using symmetry conditions, four cubic ISOBEAM elements are used to model one half of the structure. The apex load is applied in seventeen equal load steps and the restart option is employed after the first eight steps. Stiffness reformations and BFGS equilibrium iterations are employed for each load step.

Solution Results The input data on page B21.4 is used for the first eight load steps and the input data on page B21.5 is used in the restart finite element analysis.

The predicted apex displacement in the analysis is in good agreement with results reported in [ 1].

The top figure on page B21.3 shows the deformed mesh at load step 8 and load step 17. The left bottom figure on page B21.3 shows the mid-span deflection history and the reaction moment history at the build-in end.

The right bottom figure shows the history of the element axial stress closest to the mid-span and the history of the element shear stress closest to the built-in end.

User Hints

" Note that few changes are required in the input data for a restart solution and that the SOLVIA restart file, SOLVIA08.DAT, is always saved after an analysis, and available for a restart analysis.

" Results at all time steps are available in the SOLVIA-POST database since DATABASE RESTART causes the restart results to be added to the SOLVIA-POST database with results from the first run.

Version 99.0 Nonlinear Examples B21.1

SOLVIA Verification Manual Nonlinear Examples Reference

[1]

Bathe, K.J., and Bolourchi, S., Large Displacement Analysis of Three-Dimensional Beam Structures", Int. J. Num. Meth. Eng.. Vol. 14, 1979, pp. 961-986.

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SOLVIA Verification Manual Nonlinear Examples B21 LARGE DEFLECTION ANALYSIS OF A SHALLOW ARCH OR:GINAL

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SOLVIA Verification Manual Nonlinear Examples SOLVIA-PRE input HEADING

'B21 LARGE DEFLECTION ANALYSIS OF A SHALLOW ARCH' First analysis DATABASE CREATE MASTER IDOF=i00011 MODEX=EXECUTE NSTEP=8 TSTART=0 KINEMATICS DISPLACEMENT=LARGE ITERATION METHOD=BFGS TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

0. 0. /

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SYSTEM I CYLINDRICAL COORDINATES

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133.114 90.

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2 133.114 82.6603 3

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ELASTIC E=1.E7 NU=0.2 EGROUP 1

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B21A LARGE Restart analysis DATABASE OPEN MASTER MODEX=RESTART SOLVIA END SOLVIA-POST input DEFLECTION ANALYSIS OF A SHALLOW ARCH' NSTEP=9 TSTART=8 B21A LARGE DEFLECTION ANALYSIS OF A SHALLOW ARCH Restart analysis DATABASE RESTART WRITE FILENAME-'b2la.lis' SET NSYMBOLS=MYNODES VIEW=X SMOOTHNESS=YES SUBFRAME 12 MESH VECTOR=LOAD ORIGINAL=YES TIME=8 MESH VECTOR-REACTION ORIGINAL=YES SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 NHISTORY NODE=-

DIRECTION=3 OUTPUT=ALL SYMBOL=l NHISTORY NODE=2 DIRECTION=4 KIND=REACTION OUTPUT=ALL SYMBOL=2 SUBFRAME EHISTORY EHISTORY SUMMATION SUMMATION END 12 ELEMENT=l POINT=l1l KIND=SRR ELEMENT=4 POINT=341 KIND=SRS KIND=REACTION KIND=LOAD Version 99.0 Nonlinear Examples 1321.5

SOLVIA Verification Manual Nonlinear Examples EXAMPLE B22 EIGHT-STORY BUILDING SUBJECTED TO IMPACT LOAD Objective To demonstrate the capability of performing nonlinear dynamic analysis using the substructure option.

Physical Problem The eight-story building shown in figure below is analyzed for its dynamic response when subjected at the top to a triangular impact load of 30 kips-sec over 0.2 sec.

p A 200 BEAM ELEMENT BEAM ELEMENTS A

A B

SUBSTRUC TURE BEAM ELEMENT 96' SUBSTRUC TURE 15I 15' Substructure Model Beam elements:

Elastic trusses:

Elastic-plastic trusses:

ELASTIC

,PLASTIC TRUSSES ELASTIC

""TRUSSES Main Structure Model E = 1.09-10 9 psf v = 0.2 1=4 A = 1.00 ft2 p = 5.15 slugs/ft3 E = 1.00.1010 psf p = 0. slugft3 A = I E = 1.00.I10" psf cry = 6000 psf ET=

A= 10 ft2 L=10ft P

0 Version 99.0

/3 ft4 L=lOft 0 ft2 0.00 psf slugs/ft3 11"1-'

B22.1

SOLVIA Verification Manual Finite Element Model The finite element model is shown on page B22.3. The building is modeled with BEAM elements in the master structure and the substructures. The foundation is modeled with elastic TRUSS elements in the vertical direction and with elastic-plastic TRUSS elements in the horizontal direction. The dy namic response is evaluated using the trapezoidal rule (the Newmark method with u.= 0.25 and 6 = 0.50) with a time step At = 0.02 sec. A lumped mass matrix is used in the analysis. Full Newton equilibrium iterations are employed in every solution time step.

Solution Results The input data on pages B22.5 and B22.6 is used in the finite element analysis. The bottom figure on page B22.4 shows the time history of the horizontal displacement and acceleration at points D and B, see figure on page B22. 1, as displayed by SOLVIA-POST. The deformed finite element model at time t = 0.4 sec is shown in the top figure on page B22.4.

A reference solution is obtained by modeling the building as a main structure only, see figures on page B22.7 and input data on pages B22.8 and B22.9. The results of the reference solution are shown on page B22.7 and complete agreement with the substructure results can be observed.

User Hints

Note that in nonlinear dynamic analysis the convergence tolerance in the equilibrium iterations must be sufficiently tight so that the accumulation of errors in the solution is small (compare Example B2).

Version 99.0 Nonlinear Examples B22.2

SOLVIA Verification Manual ORIGINAL SOLVIA-PRE 322 EJGHT-STORY BUILDING SUBdEC-ED TO IMPACT LOAD "0.

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'B22 EIGHT-STORY BUILDING SUBJECTED TO IMPACT LOAD' DATABASE CREATE MASTER IDOF=001110 NSTEP=20 DT=0.020 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=1.E-6 TIMEFUNCTION 1

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9 0 96

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MATERIAL 2

ELASTIC E=1.09E9 NU=0.2 DENSITY=5.15 MATERIAL 3

PLASTIC ISOTROPIC E=1.E10 YIELD=6000.

ET=0.

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MESH ZONENAME:REUSEI SUBFRAME=21 MESH ZONENAME=REUSE2 SOLVIA END Version 99.0 Nonlinear Examples B22.5

SOLVIA Verification Manual SOLVIA-POST input B22 EIGHT-STORY BUILDING SUBJECTED TO IMPACT LOAD DATABASE CREATE WRITE FILENAME='b22.1is' SET VIEW=Z MESH ORIGINAL-DASHED SUBFRAME=21 SUBSTRUCTURE 1

MESH ZONENAME=REUSEl GSCALE=OLD ORIGINAL=DASHED AXIS 1 VMIN=-200 VMAX=300 LABEL='ACCELERATION' AXIS 2 VMIN=-0.6 VMAX:0.6 LABEL='DISPLACEMENT' AXIS 3 VMIN=0 VMAX=0.4 LABEL='TIME' SUBSTRUCTURE 0 NHISTORY NODE=-0 DIRECTION=i KIND=DISPLACEMENT SYMBOL=i OUTPUT=ALL SUBFRAME=21 SUBSTRUCTURE 1

REUSE 1

NHISTORY NODE=-0 DIRECTION=i KIND=DISPLACEMENT SYMBOL=2 OUTPUT=ALL SUBFRAME=OLD SUBSTRUCTURE 0

NHISTORY NODE=-0 DIRECTION=i KIND=ACCELERATION SYMBOL-i OUTPUT=ALL SUBFRAME=NEXT SUBSTRUCTURE 1

REUSE 1

NHISTORY NODE=-0 DIRECTION=I KIND=ACCELERATION SYMBOL-2 OUTPUT=ALL SUBFRAME=OLD YAXIS=2 XAXIS=3, YAXIS=-2 XAXIS=-3, YAXIS=I XAXIS=3, YAXIS=-l XAXIS=-3, NLIST END Version 99.0 Nonlinear Examples B22.6

SOLVIA Verification Manual Nonlinear Examples B22A EIGhT-S7TRY BdTLD.NG SUBJECTED TO TMPACT L'AD, REF.

ORP INAL Y

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'B22A EIGHT-STORY BUILDING SUBJECTED TO IMPACT

LOAD, REF.

SOLUTION' DATABASE CREATE MASTER IDOF=001110 NSTEP=20 DT=0.020 ANALYSIS TYPE=DYNAMIC MASSMATRIX=LUMPED METHOD=NEWMARK ITERATION METHOD=FULL-NEWTON LINE-SEARCH-YES TOLERANCES ETOL=i.E-6 TIMEFUNCTION 1

/

0.

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0.1 3.E5

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0.2 0.

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1. 0.

COORDINATES 1

0

-10

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2 15 -10

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4 0

0

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5 15 6

25 0

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8 15 48

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9 0

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10 15 96

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11 0 12 12 15 12 MATERIAL 1

ELASTIC E=I.E10 NU=0.

MATERIAL 2

ELASTIC E=1.09E9 NU-0.2 DENSITY=5.15 MATERIAL 3

PLASTIC ISOTROPIC E1=.E10 YIELD=6000.

ET=0.

EGROUP 1

TRUSS MATERIAL=i ENODES

/

1 1 4

/

2 2 5 EDATA

/

1

10.

EGROUP 2

TRUSS MATERIAL=3 ENODES

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1 3 4

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2 5 6 EDATA

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1

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BEAM MATERIAL=2 RESULTS=FORCES SECTION 1

GENERAL RINERTIA=1 SINERTIA=1 TINERTIA=1.333333 AREA=i ENODES

/

1 12 4 ii

/

2 4 ii 12 /

3 11 12 5 FIXBOUNDARIES

/

1 2 3 6 LOADS CONCENTRATED

/

9 1

0.342020

/

9 2 -0.939693 TRANSLATE EG3 Y=12 COPIES=7 SOLVIA END Version 99.0 Nonlinear Examples 1322.8

SOLVIA Verification Manual SOLVIA-POST input B22A EIGHT-STORY BUILDING SUBJECTED TO IMPACT LOAD, REF.

SOLUTION DATABASE CREATE WRITE FILENAME='b22a.lis' SET VIEW=Z MESH VECTOR=LOAD NNUMBERS=MYNODES NSYMBOL=MYNODES ORIGINAL=DASHED TIME=0.1 SUBFRAME=21 MESH ORIGINAL=DASHED AXIS 1 VMIN=-200 VMAX=300 LABEL='ACCELERATION' AXIS 2 VMIN=-0.6 VMAX=0.6 LABEL='DISPLACEMENT' AXIS 3 VMIN=0 VMAX=0.4 LABEL='TIME' NHISTORY NODE=-0 DIRECTION=i KIND-DISPLACEMENT YAXIS=2 XAXIS=3, SYMBOL=i OUTPUT=ALL SUBFRAME=21 NHISTORY NODE=8 DIRECTION=I KIND=DISPLACEMENT YAXIS=-2 XAXIS=-3, SYMBOL=2 OUTPUT=ALL SUBFRAME=OLD NHISTORY NODE=i0 DIRECTION=i KIND=ACCELERATION YAXIS=i XAXIS=3, SYMBOL=i OUTPUT=ALL SUBFRAME=NEXT NHISTORY NODE=8 DIRECTION=i KIND=ACCELERATION YAXIS=-i XAXIS=-3, SYMBOL=2 OUTPUT=ALL SUBFRAME=OLD END Version 99.0 Nonlinear Examples 1322.9

SOLVIA Verification Manual Nonlinear Examples EXAMPLE B23 ANALYSIS OF A SAND SPECIMEN UNDER COMPRESSION Objective To verify the Drucker-Prager material model when used for the PLANE AXISYMMETRIC element.

Physical Problem A uniaxial strain test for McCormick Ranch Sand is considered, see figure below.

z Young's modulus, E 100.0 ksi Poisson's ratio, v 0.25 Yield function parameter, (x 0.05 Yield function parameter, k 0.1 ksi Cap hardening parameter, W -0.066 Cap hardening parameter, D -0.78 ksi-t

\\...

\\

Tension cut-off limit, T 0.01 ksi

.Initial cap position, '1 0.0 ksi D = 1.0 in

"" H= 0.5 in Finite Element Model Four 4-node PLANE AXISYMMETRIC elements are used to model the sand specimen, see the figure on page B23.2. The top nodes in the model are subjected to prescribed displacements and the side nodes can slide in the Z-direction only. The bottom nodes are fixed.

Solution Results The input data is shown on page B23.4. One hundred prescribed displacement steps are used with the default iteration method (BFGS) and with default tolerances.

The calculated longitudinal stress as a function of the longitudinal strain is shown in the top figure on page B23.3. The calculated solution is in good agreement with the experimental data given in [1].

The traced solution path in a Drucker-Prager yield function diagram with axes -I1 and J2 is shown in the bottom figure on page B23.3. I, is the first stress invariant and J-, is the second deviatoric stress invariant, see also page 9.27 in the SOLVIA-PRE Users Manual.

We note that upon loading the first solution point represents yielding on the cap while the subsequent yielding occurs both on the cap and on the Drucker-Prager yield surface (the vertex). The cap position is moved from I1 = 0 to I1 = -2.26 during yielding.

Version 99.0 B23.1

SOLVIA Verification Manual The unloading portion is first elastic but then follows yielding on the Drucker-Prager yield surface until the tensile limit I, = T, the tensile cut-off limit, is reached.

User Hints

" For the present example one 4-node element would be satisfactory, since there is no radial displacement variation and only a linear displacement variation in the vertical (Z) direction.

" In SOLVIA, compressive stresses and strains are negative. Therefore, the cap parameters must satisfy the following conditions:

W<O D<O Ia <

T>O cc>O k>O

"* When c = 0 the Drucker-Prager yield criterion is the same as the von Mises yield criterion. For this case the yield function parameter k is the yield stress in shear and k = ay / -3, where ay is the yield stress in tension.

"* When 'I, T, i.e. the first stress invariant (the hydrostatic tension) reaches or exceeds the tension cut-off limit the deviatoric stresses and the normal stresses are set to Sij = 0 t i j = 5 ij T / 3, respectively.

Reference

[1]

DiMaggio, F.L., Sandler, I.S., "Material Model for Granular Soils", J. Eng. Mech. Div. ASCE 97 (EMS), pp. 935-949, 1971.

B23 ANALYSIS OF SAND SPECIMEN UNDER COMPRESSION ORIGINAL I

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A-RE 99 z

R,3 L_ Y IA3 R

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SQL VIA Verification Manual Nonlinear Examples B23 ANALYSIS CF SAND SPECYEjN

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SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B23 ANALYSIS OF SAND SPECIMEN UNDER COMPRESSION' DATABASE CREATE MASTER IDOF=i00111 NSTEP=100 DT=0.2 TIMEFUNCTION 1

0.

/

10. 0.033

/

20.

0.025 COORDINATES

/

ENTRIES NODE Y

Z 1

/

2 0.5

/

3 0.5 0.5 TO 5

0.

0.5 MATERIAL 1

DRUCKER-PRAGER E=100 NU=0.25 ALFA=0.05 K=0.1, W=-0.066 D=-0.78 T=0.01 I1-0.0 EGROUP 1

PLANE AXISYMMETRIC GSURFACE 1 2 3 5 ELi=2 EL2=2 NODES=4 FIXBOUNDARIES 23 INPUT=LINES 1 2 FIXBOUNDARIES 2

INPUT=LINES

/

1 5 LOADS DISPLACEMENT 3 3

-1.

TO 5 3

-1.

MESH NSYMBOLS=MYNODES NNUMBERS=MYNODES BCODE=ALL ENUMBERS=YES SOLVIA END SOLVIA-POST input B23 ANALYSIS OF SAND SPECIMEN UNDER COMPRESSION DATABASE CREATE WRITE FILENAME='b23.1is' EXYPLOT ELEMENT=1 POINT=i XKIND=EZZ YKIND=SZZ SYMBOL=1 OUTPUT=ALL EVARIABLE SIX TYPE=PLANE KIND=SXX EVARIABLE STY TYPE=PLANE KIND=SYY EVARIABLE SZZ TYPE=PLANE KIND-SZZ EVARIABLE SYS TYPE-PLANE KIND SYZ CONSTANT THREE 3.0 CONSTANT ONE 1.0 CONSTANT EPS i.E-16 RESULTANT ROOT-J2

'SQRT((SXX*SXX+SYY*SYY+SZZ*SZZ

-SXX*SYY-SYY*SZZ-SZZ*SXX+THREE*SYZ*SYZ)/THREE+EPS)'

RESULTANT

-Ii '-(SXX+SYY+SZZ)'

EPOINT ElPI EL=i POINT=i RXYPLOT XPOINT=EiPi XRESULTANT=-II YPOINT=EiPi YRESULTANT=ROOT-J2, SYMBOL=1 OUTPUT=ALL END Version 99.0 Nonlinear Examples B23*.4

SOLVIA Verification Manual EXAMPLE B24 CANTILEVER UNDER LARGE DISPLACEMENTS, SHELL Objective To verify the large displacement/rotation behaviour of the SHELL element in pure bending.

Physical Problem A cantilever beam subjected to a concentrated end moment, as shown in figure below, is considered.

L b = 1.0 in h = 1.0 in L = 12.0 in M/2 E = 1800 lb/in 2 V=0.0 M = 35 lb-in Finite Element Model The finite element model consists of one 8-node cubic SHELL element, as shown in figure on page B24.2. The load is applied in ten equal load steps in a large displacement analysis with full Newton iteration and line search. The iteration tolerance is ETOL = 10-6. The element stiffness matrix is evaluated using 4x2x2 Gauss integration.

Solution Results The input data on pages B24.4 and B24.5 is used in the finite element analysis and gives the following results for the tip rotation p/27c and the deflection ratios u/L and w/L:

M u

w 2n L

L Theory SOLVIA Theory SOLVIA Theory SOLVIA 7.0 0.089 0.089 0.051 0.052 0.273 0.273 14.0 0.178 0.179 0.196 0.196 0.504 0.505 21.0 0.267 0.269 0.408 0.408 0.660 0.667 28.0 0.357 0.364 0.650 0.662 0.723 0.744 35.0 0.446 0.496 0.880 0.991 0.694 0.704 Version 99.0 z

Y M/2 X

Nonlinear Examples B24.1

SOLVIA Verification Manual Nonlinear Examples The theoretical values assuming small strains and thin shell behaviour are calculated using ML 2nt 27tEI u

I1 E l s in MI L

L M

EI) w El (ML L

ML 1 EI-

)

where I is the moment of inertia (bh' / 12).

The top figure on page B24.3 shows the deformed finite element meshes at M=14 and M=35. The displacement ratios u/L, w/L and 4/27r as a function of the load are shown in the bottom figure on page B24.3.

User Hints "Note that the predicted response in the finite element analysis compares well with the analytical solution up to approximately 90 degrees rotation of the cantilever tip, but for larger rotations the model is to coarse to define the geometry of the deformed cantilever accurately. As an example of the geometrical discretization error, the figure on page B24.4 shows the midsurface of a cubic shell element for which the nodes lie on a half circle ( =

", For rotations in excess of about 90 degrees more elements need, therefore, be employed.

B24 CANTILEVER UNDER LARGE DISPLACEMENTS.

SHELL ORIGINAL.

TiME I

z X

y

.S MOMENT

OLVTA-PR 99 2 C0LVIA ENGINE-RNGC AB Version 99.0 B24.2

SOLVIA Verification Manual Nonlinear Examples 324 CAN77L EVE HONER L ARGE 0SPLACE,>1EN

--- H

023 0 LVIA-POS3 99.0 SOLVIA ENGINEERING AB B2A CANTILEVER UNDER L.ARGE DISPLACEMENTS SHEL 0

1

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'MJHN' jN SQ VII ENO.N'E-RNG AB Version 99.0B4.

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-AI 2ý SCLV:A-P0S7 99A B24.3

SOLVIA Verification Manual 10-0.5 0

0.5 1.0 SOLVIA-PRE input HEADING

'B24 CANTILEVER UNDER LARGE DISPLACEMENTS, SHELL' DATABASE CREATE MASTER IDOF=010101 NSTEP=i0 DT=i.

KINEMATICS DISPLACEMENT-LARGE ITERATION METHOD=FULL-NEWTON LINE-SEARCH=YES TOLERANCES ETOL=i.E-6 TIMEFUNCTION 1

0.

0. /

10.

35.

COORDINATES 1

TO 4

12.

/

5

0.

1.

TO 8

12.

1.

MATERIAL 1

ELASTIC E=i800.

NU=-.

EGROUP 1

SHELL SINT=2 RESULTS=STRESSES ENODES ENTRIES EL N1 N2 N3 N4 N5 N7 N9 Nil 1

1 4

8 5

2 7

3 6

THICKNESS 1

1.

FIXBOUNDARIES

/

1 5 LOADS CONCENTRATED 4

5

-0.5 8 5 -0.5 MESH NSYMBOLS=YES NNUMBERS=YES VECTOR=MLOAD SOLVIA END Version 99.0 Nonlinear Examples 1324.4

SOLVIA Verification Manual SOLVIA-POST input B24 CANTILEVER UNDER LARGE DISPLACEMENTS, SHELL DATABASE CREATE WRITE FILENAME:'b24.1is' TOLERANCES SET SMOOTHNESS=YES HIDDEN-REMOVAL=NO COLOR MODE=LINE MESH TIME=4 ORIGINAL=DASHED MESH TIME=-0 GSCALE=OLD SUBFRAME-OLD TEXT=NO NPOINT NAME=TIP NODE=4 NVARIABLE NAME=U DIRECTION=I KIND=DISPLACEMENT NVARIABLE NAME=W DIRECTION=3 KIND=DISPLACEMENT NVARIABLE NAME=PHI DIRECTION=5 KIND-DISPLACEMENT CONSTANT NAME=L VALUE=I2.

CONSTANT NAME-TWOPI VALUE=6.2832 RESULTANT NAME=UOVERL STRING='-U/L' RESULTANT NAME=WOVERL STRING='W/L' RESULTANT NAME=ANGLE STRING='-PHI/TWOPI' AXIS ID-i VMIN=-

VMAX=i LABEL='DISP RATIOS' RHISTORY POINTNAME=TIP RESULTANTNAME=UOVERL XVARIABLE=i YAXIS=i, OUTPUT=ALL SYMBOL=i RHISTORY POINTNAME=TIP RESULTANTNAME=WOVERL XVARIABLE=i YAXIS=-i, OUTPUT=ALL SYMBOL=2 SUBFRAME=OLD RHISTORY POINTNAME=TIP RESULTANTNAME=ANGLE XVARIABLE=i YAXIS=-i, OUTPUT=ALL SYMBOL=3 SUBFRAME=OLD END Version 99.0 Nonlinear Examples B24.5

SOLVIA Verification Manual EXAMPLE B25 CANTILEVER UNDER LARGE DISPLACEMENTS, PLATE Objective To verify the large displacement/rotation behaviour of the PLATE element in pure bending.

Physical Problem The same type of problem as in Example B24 is considered. The dimensions of the cantilever and the Young's modulus of the material are, however, different as can be seen in the figure below.

z I -y-M12 X

L b= 10.0in E = 12000 lbin2 h= 0.5 in L = 100.0 in v=0.0 M = 35 lb-in Finite Element Model The finite element model considered is shown in the figure on page B25.2. The model consists of ten PLATE elements and the end moment is applied in ten equal load steps in a large displacement analysis.

The solution is obtained using the full Newton iteration. The iteration tolerances are RTOL = 0.01, RNORM = 0.35 and RMNORM = 35.

Solution Results The input data on pages B25.4 and B25.5 gives the following results in the finite element analysis:

M u

w 27c L

L Theory SOLVIA Theory SOLVIA Theory SOLVIA 7.0 0.089 0.089 0.051 0.052 0.273 0.275 14.0 0.178 0.178 0.196 0.197 0.504 0.507 21.0 0.267 0.267 0.408 0.411 0.660 0.664 28.0 0.357 0.357 0.650 0.654 0.723 0.728 35.0 0.446 0.446 0.880 0.885 0.694 0.698 The theoretical results are calculated as in Example B24.

Version 99.0 I0 Nonlinear Examples B25.1

SOLVIA Verification Manual The top figure on page B25.3 shows the deformed finite element meshes at M = 14 and M = 35. The bottom figure on page B25.3 shows the displacement ratios 4/2r. u/L and w!L as a function of the load as displayed by SOLVIA-POST.

User Hints

° Since more elements are used in this example, a better solution is obtained than in Example B24, where only one cubic SHELL element is employed.

B25 CANT LEVER UNDER LARGE DT SPLAUEMENTS, PLATE OR TS:NAL S.

TIME I

z 7

Yr MOMENT 1.75 IAS-ER QVA A RN tl SOLVIA ENG7NEERTNIG AB rOLVIA-PRE 99.0 Version 99.0 I

Nonlinear Examples

ý I B25.2

SOLVIA Verification Manual 32S CANTILEVER UNDER

-ARGE DISPLACEMENTS OR

IJNAL 1"

1.

MAX DISPL 54 126 TiME 4 Nonlinear Examples PLATE z

X y

SOL',/IA-POST 99.0 SOLVIA ENGINEERING AB B2S CANTILEVER UNDER LARGE DISPLACEMENTS, PLATE

-U

.+-

...1.

19 1 S 2S VALUE DF T MEF iNCAON t

COL/IA-2CST 99.

L\\/IA ENG\\NEERING AB Version 99.0 jF B25.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B25 CANTILEVER UNDER LARGE DISPLACEMENTS, PLATE' DATABASE CREATE MASTER IDOF=010101 NSTEP=10 DT=1 KINEMATICS DISPLACEMENTS=LARGE ITERATION METHOD=FULL-NEWTON TOLERANCES TYPE=F RTOL=0.01 RNORM=.35 RMNORM=35 TIMEFUNCTION 1

0.

/

10. 35.

COORDINATES 1

TO 6

100.

7

0.

10.

TO 12 100.

10.

MATERIAL 1

ELASTIC E=12000.

NU=0.

EGROUP 1

PLATE ENODES 1

1 2 7 TO 5

5 6 11 6

2 8 7 TO 10 6 12 11 EDATA

/

1

.5 FIXBOUNDARIES

/

1 7 LOADS CONCENTRATED 6 5 -0.5 12 5

-0.5 MESH NSYMBOLS=YES NNUMBERS=YES VECTOR=MLOAD BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B25.4

SOLVIA Verification Manual SOLVIA-POST input B25 CANTILEVER UNDER LARGE DISPLACEMENTS, PLATE DATABASE CREATE TOLERANCES WRITE FILENAME='b25.1is' MESH ORIGINAL=DASHED TIME=4 MESH GSCALE=OLD SUBFRAME=OLD TEXT=NO NPOINT NAME=TIP NODE=6 NVARIABLE NAME=U DIRECTION=i KIND=DISPLACEMENT NVARIABLE NAME=W DIRECTION=3 KIND=DISPLACEMENT NVARIABLE NAME=PHI DIRECTION=5 KIND=DISPLACEMENT CONSTANT NAME=L VALUE=I00.

CONSTANT NAME=TWOPI VALUE=6.2832 RESULTANT NAME=UOVERL STRING='-U/L' RESULTANT NAME=WOVERL STRING='W/L' RESULTANT NAME=ANGLE STRING='-PHI/TWOPI' AXIS ID=i VMIN=O VMAX=i LABEL-'DISP RATIOS' RHISTORY POINTNAME=TIP RESULTANTNAME=UOVERL XVARIABLE-i SYMBOL=-

OUTPUT=ALL RHISTORY POINTNAME=TIP RESULTANTNAME=WOVERL XVARIABLE-i SYMBOL=2 OUTPUT=ALL SUBFRAME=OLD RHISTORY POINTNAME=TIP RESULTANTNAME=ANGLE XVARIABLE=i SYMBOL=3 OUTPUT=ALL SUBFRAME=OLD END Version 99.0 YAXIS=i, YAXIS=-i, YAXIS=-i, Nonlinear Examples 1325.5

SOLVIA Verification Manual EXAMPLE B26 PLASTIC CANTILEVER UNDER PURE BENDING, SHELL Objective To verify the elastic-plastic behaviour of the SHELL element.

Physical Problem The cantilever beam subjected to a concentrated end moment, as shown in the figure below, is considered. The elastic-plastic response of the cantilever is to be determined.

z L

X T.

E M/2 Y

M12 b=2.0 in h=.0 in L=12.0 in E =3.0.10' psi v= 0.3 Gy = 1.5.103 psi ET =3.0.10' psi M = 900 lb-in Finite Element Model Due to symmetry conditions, only one half of the cantilever is modeled. The finite element model consists of one 16-node SHELL element as shown in the figure on page B26.2. The material of the cantilever is modeled as elastic-plastic with isotropic hardening. The element stiffness matrix is evaluated using 4x4x6 Gauss integration. The solution response is traced with the AUTOSTEP method using BFGS iteration. The maximum solution time step is set to 3 and request is made of intermediate solutions at 6 specified solution times.

Solution Results The input data on pages B26.4 and B26.5 is used in the finite element analysis.

In the table on the next page the calculated tip rotations for 4x4x6 and 4x4x4 Gauss integration are compared with an analytical solution using beam theory.

Version 99.0 Nonlinear Examples B26.1

SOLVIA Verification Manual The figures on page B26.3 show the variation of the applied moment versus the tip rotation as pre dicted in the finite element analysis with 4x4x6 and 4x4x4 Gauss integration. For comparison, the analytical beam theory solution is also presented in the figures.

User Hints

"° Note that in an elastic-plastic analysis the order of Gauss integration through the element thickness must be high enough to predict the plastic behaviour through the shell thickness accurately.

Note that the anticlastic bending effect gives a slight variation of the transverse displacement along the width of the shell element.

B26 PLASTIC CANTILEVER UNDER PURE BENDING.

SHELL, TINT=6 ORIGINAL I

Z TIrE 3 x

Y 7-EAXESRST 4

b~IA TE C Il C D

10 0 C C

OLVIA ENGTNEER.AING AB SOL!IA-PRE 99,0 Version 99.0 Rotation, qb Moment Beam theory SOLVIA SOLVIA (4x4x6)

(4x4x4) 500 1.20.10-.

1.20.10-'

1.20.10.3 600 1.55 10. 3 1.56-10-3 1.59.10-3 700 2.59.10.'

2.88.10.3 2.59.10-3 800 1.42.10-2 1.03.10-2 6.28.10-3 900 3.79.10-2 3.42.10-2 3.02-10-2 Nonlinear Examples B26.2

SOLVIA Verification Manual Nonlinear Examples B26 PLASTIC CAN71LEVER UNDER PURE BENDING,

SHELL, TINT=6 0.0 5.0 01)'.

1 5.0 20.0 25.0 PHI SOLVIA-POST 99.0

33.

0 30 SOLVIA ENGINEERING AB B26A PLASTIC CANTILEVER UNDER PURE BENDING,

SHELL, TINT=-

A z

250 30.0 SOLVIA

\\NG NEERING AB Version 99.0 3

0 z

o (

70

C' 0'

z H'

13 0

10.0 SOLV:A-POST

)9.0 20 PHI o-ra !/

B26.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B26 PLASTIC CANTILEVER UNDER PURE BENDING,

SHELL, TINT=6' DATABASE CREATE MASTER NSTEP=20 DT=3.0 ITERATION METHOD=BFGS TOLERANCES ETOL=1.E-8 AUTO-STEP DTMAX 3.0 TMAX 18 T=I10 12 13 14 15 16 TIMEFUNCTION 1

0. 0. /

18.

1.8 COORDINATES 1

/

2 1.0

/

3 1.0 12.

/

4

0.

12.

MATERIAL 1

PLASTIC E=3.E7 NU=0.3 YIELD-1.5E3 ET=3.E5 EGROUP 1

SHELL TINT=6 THICKNESS 1

1.

GSURFACE 1 2 3 4 EL1=1 EL2=1 NODESf16 LOADS ELEMENT TYPE=MOMENT INPUT=LINE 3 4 edge 250.

250.

FIXBOUNDARIES 246 INPUT*LINE 1 2 FIXBOUNDARIES 156 INPUT=LINE

/

1 4 FIXBOUNDARIES 3

INPUT-NODE

/

1 FIXBOUNDARIES 6

INPUT=LINE

/

3 4 MESH NSYMBOLS=YES NNUMBERS=MYNODES VECTOR=LOAD BCODE=ALL, EAXES=RST SOLVIA END Version 99.0 Nonlinear Examples 1326.4

SOLVIA Verification Manual SOLVIA-POST input B26 PLASTIC CANTILEVER UNDER PURE BENDING, SHELL DATABASE CREATE WRITE FILENAME='b26.1is' AXIS 1 VMIN=O.

VMAX=4.OE-2 LABELSTRING='PHI' AXIS 2 VMIN=O.

VMAX=900 LABELSTRING='BENDING MOMENT' NVARIABLE NAME=XMOM DIRECTION=4 KIND=REACTION NVARIANLE NAME=XROT DIRECTION=4 KIND=DISPLACEMENT CONSTANT NAME=TWO VALUE=-2.0 CONSTANT NAME=EIGHT VALUE=8.0 RESULTANT NAME=PHI

'XROT' RESULTANT NAME=MOMENT

'XMOM*TWO*EIGHT' NPOINT NAME=N1 NODE=I NPOINT NAME=N4 NODE=4 RXYPLOT XPOINT=N4 XRESULTANT=PHI YPOINT=NI YRESULTANT=MOMENT, SYMBOL=i TSTART=3 XAXIS=i YAXIS=2 OUTPUT=ALL USERCURVE 1 READ B26.DAT PLOT USERCURVE 1 XAXIS=-I YAXIS=-2 SUBFRAME=OLD END Version 99.0 Nonlinear Examples B26.5

SOLVIA Verification Manual EXAMPLE B27 PLASTIC CANTILEVER UNDER PURE BENDING, PLATE Objective To verify the elastic-plastic behaviour of the PLATE element.

Physical Problem Same as in figure on page B26-1.

Finite Element Model The finite element model considered is shown in the figure on page B27.2. Due to symmetry only one half of the cantilever beam is modeled using eight PLATE elements. The material of the beam is modeled as elastic-plastic with isotropic hardening and Ilyushin yield condition. The solution re sponse is traced with the AUTOSTEP method using BFGS iteration. The maximum solution time step is set to 3 and request is made of intermediate solutions at 6 specified solution times.

Solution Results The input data on pages B27.3 and B27.4 is used in the finite element analysis.

The following results for the tip rotation 0 (rad) are obtained in the analysis:

Moment SOLVIA Beam theory 500 1.20.10-3 1.20.10-'

750 1.80.10-3 5.10 -10 900 4.97.10-2 3.79.10-2 The bottom figure on page B27.2 shows the variation of the applied moment on the whole plate versus the tip rotation as predicted in the finite element analysis. An analytical beam theory solution is also presented in the same figure.

User Hints

" The PLATE element stiffness matrix is calculated using stress resultants and no numerical integra tion in the thickness direction is used. The yield condition employed is the one of Ilyushin [1], and the stress resultants conrespond either to elastic or plastic conditions. This means that the entire cross section at an integration point changes at the same time from elastic to plastic and vice versa.

" In this example with pure moment loading the plastic response is calculated based on the condi tions at the distance h/4 from the midsurface, where h is the thickness. However, in the case of strain-hardening the incremental plastic moment can be represented by two force resultants each located h/3 from the midsurface, which is the position of the center of gravity of the corresponding triangular incremental stress distribution. Hence, the calculated plastic slope in the moment rotation diagram on page B27.2 is 3/4 of the theoretical plastic slope asymptote.

Version 99.0 Nonlinear Examples B27.1

SOLVIA Verification Manual Nonlinear Examples 327 PLASTIC CANTILEVER UNDER PURE BENDING, PLATE TRI: ENA T'IE X

IA S P,

B C'101.

C 0o0 1 D 1ii':

O SOLVIA ENGINEERING AB SOLVIA-PRE 99.0 B27 PLAS7IC CANTILEVER UNDER PURE BENDING, PLATE 0

'I) o~~ ~~~~~

~~~~

........i............

S1

,-. ).

20.r0 2.

PHL SOkVT A-POST 99.8 S. C 3S.

40. 3

'15.C s0 SCLVIA ENGINEERING AB Version 99.0 0

1327.2

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B27 PLASTIC CANTILEVER UNDER PURE BENDING, PLATE' DATABASE CREATE MASTER NSTEP=20 DT=3.0 ITERATION METHOD=BFGS TOLERANCES ETOL=1.E-6 AUTO-STEP DTMAX=3.0 TMAX=18.

T1=10 12 13 14 15 16 TIMEFUNCTION 1

0. 0.

/

18.

1.8 COORDINATES 1

TO 5

0.

12.

/

6 1.0 TO 10 1.0 12.

MATERIAL 1

ILYUSHIN E=3.E7 NU=0.3 YIELD=1.5E3 ET=3.E5 EGROUP 1

PLATE ENODES 1 1 2 6

/

2 2 3 8

/

3 3 4 8

/

4 4 5 10 5

6 2 7

/

6 2 8 7

/

7 8

4 9

/

8 4 10 9 EDATA

/

1

1.

FIXBOUNDARIES 15

/

2 TO 5 FIXBOUNDARIES 12345

/

1 FIXBOUNDARIES 24

/

6 LOADS CONCENTRATED 5 4 125.

10 4 125.

MESH NSYMBOLS=YES NNUMBERS=YES VECTOR=MLOAD BCODE=ALL SOLVIA END Version 99.0 Nonlinear Examples B27.3

SOLVIA Verification Manual SOLVIA-POST input B27 PLASTIC CANTILEVER UNDER PURE BENDING, PLATE DATABASE CREATE WRITE FILENAME='b27.1is' AXIS 1 VMIN=0.

VMAX=5.OE-2 LABELSTRING-'PHI' AXIS 2 VMIN=0.

VMAX:900 LABELSTRING='BENDING MOMENT' NVARIABLE NAME=XMOM DIRECTION=4 KIND=REACION NVARIANLE NAME=XROT DIRECTION=4 KIND=DISPLACEMENT CONSTANT NAME=TWO VALUE=2.0 RESULTANT NAME-PHI

'XROT' RESULTANT NAME=MOMENT XMOM

TWO

TWO' NPOINT NAME=N1 NODE=i NPOINT NAME=N5 NODE=5 RXYPLOT XPOINT=N5 XRESULTANT=PHI YPOINT=Ni YRESULTANT=MOMENT, SYMBOL=i TSTART=3 XAXIS=i YAXIS=2 OUTPUT=ALL USERCURVE 1 READ B26.DAT PLOT USERCURVE 1 XAXIS=-i YAXIS=-2 SUBFRAME=OLD END Version 99.0 Nonlinear Examples 1327.4

SOLVIA Verification Manual EXAMPLE B28 TRANSIENT TEMPERATURE ANALYSIS OF A SLAB Objective To verify the behaviour of the TRUSS conduction element in SOLVIA-TEMP when subjected to simultaneous boundary convection and radiation in a transient analysis.

Physical Problem The slab shown in the figure below is considered. The slab is initially at a uniform temperature 0i and at time t = 0+ the slab surfaces are exposed to convection and radiation. The time history of the tem perature at the surface and at the center of the slab is to be determined.

Go SIMULTANEOUS BOUNDARY a--.

CONVECTION AND RADIATION L

L SIMULTANEOUS BOUNDARY

-?CONVECTION AND RADIATION e =Or =00 o0 Thickness/2 L = 1.0 in Convection coefficient h = 0.04 Btu/F

hr. in 2 Conductivity k=0.01 Btu/°F.hr.in Shape factor f= 1.0 Specific heat c =0.01 Btu/°F.in 3 Emissivity constant c

1.0 (black body)

Thermal diffusivity a k/ c = 1.0 in 2 / hr Bolzmann's constant o=0.118958-10-m Btu/ in 2 -hr-° R 4

-=._-vfge-9'

-L = 4.0 gives 0, = 1498.15 k

h.L Bi=-

4.0 k

Finite Element Model Because of symmetry conditions only one half of the slab is modeled. The finite element model consists of twenty equally spaced TRUSS conduction elements as shown in the left top figure on page B28.3. The thermal conductivity and the heat capacity of the material are assumed to be constant. A diagonal heat capacity matrix is used. The transient temperature response is evaluated using the Euler backward method in the step-by-step analysis. Conductivity reformation and heat flow equilibrium iteration using the modified Newton-Raphson method is performed at each time step.

Version 99.0 Nonlinear Examples B28.1

SOLVIA Verification Manual In order to obtain the required accuracy in the SOLVIA-TEMP analysis the time step size is changed during the response prediction.

t* =aC. t / L2 At* =.At / L2 Number of time steps First interval 0.00<t*<0.06 0.001 60 Second interval 0.06<t*<0.66 0.01 60 Third interval 0.66<t*<3.66 0.05 60 Solution Results The input data on pages B28.4 to B28.5 is used in the finite element analysis. The solution of the sur face temperature and the temperature at the center of the slab, 0s and 0C respectively, depends upon two parameters; the Biot number Bi and the radiation parameter F. In this analysis Bi = 4 and F = 4.

The figure below shows the temperature variation at the surface and at the center of slab for Bi = 4, and for F= 0 and F= 4. The SOLVIA-TEMP solutions are compared with solutions obtained by Haji-Sheikh and Sparrow [I] who used a probability method. The temperature variations as displayed by SOLVIA-POST are shown in the bottom figure on page B28.3.

The right top figure on page B28.3 shows the distribution of heat flux along the model at time 3.66 and the temperature distributions at times 0.03 and 0.06.

Reference

[1]

Haji-Sheikh, A., and Sparrow, E.M., "The Solution of Heat Conduction Problems by Probability Methods", Trans. ASME, J. Heat Transfer, Vol. 39, pp. 121-131, 1967.

Version 99.0 1.0 0

.8 -

.2 98

-4.0 oc 4-

4.

-. 6 HAJI-SHEIKH AND

0.

SPARROW 0 SOLVIA-TEMP SOLUTION

.2 -

.8 A SOLVIA-TEMP SOLUTION 0

1 1

8, 4.O 1

1.0 2

4 6 810-2 2

4 6 810-2 4 6 810 2

4 6 cct/L 2 Nonlinear" Examples B28.2

SOLVIA Verification Manual Nonlinear Examples 328 TRANSIENT TEMPEIFAP RE ANALYSIS OF A SLAB

-RIGINAL 3,

z

-x 2

3 41.5

.5 7

9 0o 'It 12 i3 4 t 1 6.1 18 19 20_2t ORIGINAL 3

Z L, t

2 3

4 5

6 7

8 9

10 t:2 13 i1, LS L6 '7 18 19 20 328 TRANSHNT TEMPERATURE ANALYSIS OF A SLAB TIME s6

~2 0,

A3 2,

0.4 25 6

TRUSS24-113 01 0

SOLVA

-NG:NEER.NG AB CISTANCE SOLVIA-POST 99.0 SOLVIA ENGTNEERING B28 TRANSTENVT TEPERATURE ANALYSIS CF A SLAB 30 1 10

2.

2 3 03S

\\

O1O 1

.0' 2.3 2

5 3.0 3.

1.0 50LVIA-PDST 99.0 SOLV7A ENG7\\'DR NG AB Version 99.0 SOLVIA-PRE 99

[.0 AS L-L,.

B28.3

SOLVIA Verification Manual SOLVIA-PRE input HEADING

'B28 TRANSIENT TEMPERATURE ANALYSIS OF A SLAB' DATABASE CREATE T-MASTER NSTEP=60 DT=0.001 60 0.01 60 0.05 T-ANALYSIS TRANSIENT HEATMATRIX=LUMPED METHOD=BACKWARD-EULER T-ITERATION TTOL=1.E-4 T-ITERATION METHOD=MODIFIED-NEWTON TIMEFUNCTION 1

0. 0. /

10.

0.

COORDINATES 1

TO 21

1.

T-INITIAL TEMPERATURES 1

1498.1505 TO 21 1498.1505 T-MATERIAL T-MATERIAL T-MATERIAL EGROUP ENODES EDATA 1 CONDUCTION K=0.01 SPECIFICHEAT=0.01 2 CONVECTION H=0.04 3 RADIATION EMISSIVITY=1.

SIGMA=0.118958E-10, TEMPUNIT=KELVIN 1

TRUSS MATERIAL=I

/

1 1 2 TO 20 20 21

/

1 1.

T-BOUNDARIES SEGMENTS CON-MATERIAL=2 RAD-MATERIAL=3 INPUT=NODES 1 21 1.

1.

T-LOADS ENVIRONMENTAL CONVECTION=YES RADIATION=YES INPUT-NODES 21 1.

SET MESH MESH PLOTORIENTATION=PORTRAIT VIEW=-Y NSYMBOLS=YES NNUMBERS=YES SUBFRAME=12 VIEW=-Y NSYMBOLS=YES ENUMBERS=YES SOLVIA-TEMP END Version 99.0 Nonlinear Examples B28.4

SOLVIA Verification Manual SOLVIA-POST input B28 TRANSIENT TEMPERATURE ANALYSIS OF A SLAB T-DATABASE CREATE WRITE FILENAME='b28.1is' EPLINE NAME=TRUSS

/

1 1 TO 20 1 NPLINE NAME=MEDIUM

/

1 TO 21 AXIS ID=i VMIN=0 VMAX=1 LABEL-'DISTANCE' AXIS ID=2 VMIN=-

VMAX=1500 LABEL='TEMPERATURE' SET PLOTORIENTATION=PORTRAIT SUBFRAME 12 ELINE LINENAME=TRUSS KIND=TFLUX SYMBOL=1 OUTPUT=ALL NLINE LINENAME=MEDIUM KIND=TEMPERATURE XAXIS=1 YAXIS=2, TIME=0.03 OUTPUT=ALL SYMBOL=I NLINE LINENAME=MEDIUM KIND=TEMPERATURE XAXIS=--

YAXIS=-2, TIME-0.06 OUTPUT=ALL SYMBOL=4 SUBFRAME=OLD SET PLOTORIENTATION=LANDSCAPE NVARIABLE T KIND=TEMPERATURE CONSTANT TI 1498.1505 RESULTANT T-BY-TI 'T/TI' NPOINT SURFACE NODE=21 NPOINT CENTER NODE=1 SUBFRAME 12 RHISTORY POINT=SURFACE RESULTANT=T-BY-TI RHISTORY POINT=CENTER RESULTANT=T-BY-TI NLIST END Version 99.0 Nonlinear Examples B28.5

SOLVIA Verification Manual EXAMPLE B29 TRANSIENT TEMPERATURE ANALYSIS OF A SPACE SHUTTLE Objective To verify the behaviour of the TRUSS conduction element using nonlinear material models in a transient analysis.

Physical Problem The space shuttle orbiter thermal protection system shown in the figure below is to be analyzed. The protection system is composed of different materials and the thermophysical properties of these mate rials are given on page B29.3. The uniform initial temperature is 322 K. At time t = 0+ a step heat flow input is imposed on the surface of the thermal protection system and maintained for 100 sec.

after which no heat flow input is imposed. Simultaneously, the surface of the protection system is exposed to radiation to a sink of absolute zero temperature (0, = 0 K). It is desired to predict the transient surface temperature of the protection system. The emissivity coefficient of the radiation surface is 0.85.

HEAT FLOW INPUT RADIATION HEAT FLUX 8, = O'K 1.36 x 05 q (w/m2i (D

Z 0

(-'

od 03 oo

(.D (n

H

_J LU ILi o

In 5,

H 2:

-7 TF

-1/1- ;#fl 381 cm

.01905 cm THERMAL PROTECTION SYSTEM 0

50 100 150 TIME (sec)

Finite Element Model The finite element model considered is shown in the bottom figures on next page. It consists of nineteen TRUSS conduction elements and one BOUNDARY SEGMENT for radiation. A lumped heat capacity matrix is used. Conductivity reformations and heat flow equilibrium iterations using the modified Newton-Raphson method are performed at each step of the analysis. The ct-family time integration with ct = 0.7 and with time steps as shown in the table below are used:

Version 99.0 Nonlinear Examples qS B29.1

SOLVIA Verification Manual Nonlinear Examples Solution Results The solution for the surface temperature and the net heat flow from node 1 due to conduction is shown as function of time in the top figures on page B29.4 using the input data on pages B29.4 to B29.6. This same problem was also solved by Williams and Curry who used the finite difference method [ 1] and good agreement with the SOLVIA-TEMP solution presented here can be observed.

Note that the net nodal heat flow from node 1 due to conduction is not equal to the heat flow input to node 1, which is prescribed by T-LOADS HEATFLOW. The reason is that some of the input heat flow is used to heat the material lumped at node 1 and some of the input heat flow is radiated to the environment. The net remaining heat flow from node 1 is due to conduction and is shown in the right top figure of page B29.4.

Reference

[1]

Williams, S.D., and Curry, D.M., "An Implicit-Iterative Solution of the Heat Conduction Equa tion with a Radiation Boundary Condition", Int. J. Num. Meth. Eng., Vol. 11, pp. 1605-1620, 1977.

2-1 2-2 ORIGINAL 0.0002 ZONE E33

.20 3-2 SOLVlA-PRE 99.0 SOLVIA ENGINEERING AB B29 TRANSIENT TEMPERATURE ANALYSIS OF A SPACE SHUTTLE ORIGINAL 0.0000S Z

ZONE ECI L_

4-i ORIGINAL 0.01 Z

ZONE EGS 2

i-7 9 A

!0 2 2 S -

oS5 56 S-7 S-8 59 5_1-0 ORIGINAL 0 01 SOL/IA-PRE 99.0 SOLVIA ENG:NEERING AB Version 99.0 Time (sec)

Time step DT (sec)

Number of time steps 0 < t < 2 0.05 40 2<t< 10 0.5 16 10 < t < 100 1.0 90 100 < t* 110 0.5 20 110 < t* 150 1.0 40 329 TRANSIENT TEMPERATURE ANALYSIS OF A SPACE SHUTTLE OREGINAL

-0.U005 Z

ZONE EGI L

"2 13

_7 38 1-L 1-2 ORIGINAL 0000S Z

ZONE E32 LK B29.2

SOLVIA Verification Manual Nonlinear Examples Thermophysical Properties 0 30 r 0.25 I-6 I

0 015 010

, 005 0

!2500

-. 250 o000 750 500 250 S~0 200 400 600 800 1000 1200 1400 1600 TEMPERATURE (`<)

THERMOPHYSICAL PROPERTIES OF RSJ (densiiy = 144.17 g/m3) 2500 2000 1500 1000

=

00 100 200 300 400 500 600 700 800 TEMPERATURE (*K)

THERMOPHYSICAL PROPERTIES OF FEL7 (densty =96 11 k491m3)

E z

3 3.5 3.0 2.5 2.0 1.5 1.0 0.5 0.0 K

0.4 2000 0

0

.4j 0 2 1-4600 T

K 02esa

-40 000Qm3 4500

-,500 0

0 S

0 100 200 300 400 500 600 700 800 TEMP2E44TU1CC V°Kl TEMPERATURE. ( K)

THERMOPHYSICAL PROPERTIES OFALUMINUM (denlit

= 2851.28 W91 3) 71750 1500 1250

'1000 750 500 250 Version 99.0 0.05 0 04 0.03 002 0.01 0

8 0

0 200 400 600 800 000 i200 1400 1600 1800 200 TEMPERATURE (1K)

THERMOPHYSICAL PROPERTIES OF 042 COATING (dans.ry=1665.92 k8/mf

emissivity 0

0.8) 0 B29.3

SOLVIA Verification Manual Nonlinear Examples 329 TRANSI\\NT TEMPE-RATURE ANALVS3S CF A SPACE SHUTTLE

-I a';

L N

4 7.1 77 SCLV.A-POST 99.0 SOLVIA '-NCINEW7R7,NC AB SOLVIA-PRE input HEAD

'B29 TRANSIENT TEMPERATURE ANALYSIS OF A SPACE SHUTTLE' DATABASE CREATE T-MASTER NSTEP-40 DT-O.05 16 0.5 90 1.0 20 0.5 40 1.0 T-ANALYSIS TYPE-TRANSIENT HEATMATRIX-LUMPED METHOD-ALPHA-FAMILY, ALPHA-C. 7 T-ITERATION TTOL=i.E-4 METHOD=MODIFIED-NEWTflN TIMEFUNCTION 1

0. 1.

/100.

1 /

101. 0.

/200.

0.

TIMEFUNCTION 2

0. 0.

/200.

0.

COORD INATES 1

/

2 0.000381

/

3 0.001381

/4 0.004041 5

0.011041 TO 12 0.074041 13 0.0742315

/

14 0.0746285 15 0.0756285 TO 17 0.0784225 18 0.0786130 To 20 0.0801370 T-INITIAL TEMPERATURE TREME322.22 T-MATERIAL 1 CONDUCTION K=0.3113 SPECIFICHEATD1.71037652E6 T*EUCTO Version 99.0 B29.4

Nonlinear Examples SOLVIA Verification Manual SOLVIA-PRE input (cont.)

T-MATERIAL 277.78 477.78 555.56 T-MATERIAL 111.11 166.67 222.22 277.78 333.33 388.89 444.44 T-MATERIAL 255.56 394.44 672.22 950.00 1200.00 1366.67 1533.33 2200.00 T-MATERIAI 255.56 394.44 533.33 672.22 811.11 950.00 1088.89 1200.00 1227.78 1338.89 1366.67 1422.22 1450.00 1533.33 T-MATERIAL EGROUP 1

ENODES

/

EDATA

/

EGROUP 2

ENODES

/

EDATA

/

EGROUP 3

ENODES

/

EDATA

/

EGROUP 4

ENODES

/

EDATA

/

Version 99.0 2

TEMP-CONDUCTION 0.0294 116616.03 0.0363 116616.03 0.0484 116616.03 3

TEMP-CONDUCTION 75.9611 2.743843E6 95.2627 2.743843E6 108.9606 2.743843E6 120.7906 2.743843E6 131.3753 2.743843E6 141.9601 2.743843E6 151.9222 2.743843E6 4

TEMP-CONDUCTION 0.8423 1.324340E6 0.9512 1.672850E6 1.1311 1.812254E6 1.2971 1.986510E6 1.4338 2.195616E6 1.5272 2.404722E6 1.6137 2.509275E6 1.9472 2.897035E6 L

5 TEMP-CONDUCTION 0.0476 0.0592 0.0749 0.0922 0.1138 0.1354 0.1629 0.1889 0.1960 0.2262 0.2349 0.2522 0.2608 0.2883 0.114609E6 0.129690E6 0.144770E6 0.156834E6 0.171914E6 0.180962E6

0. 190010E6

0. 196042E6

0. 199058E6

0. 205090E6 0.208106E6

0. 211122E6 0.212932E6 0.217155E6 6

RADIATION EMISSIVITY=0.

8 5 SIGMA=5.67E-8 TEMP=KELVIN TRUSS MATERIAL=i 1 12 13

/

2 17 18 1 1.

TRUSS MATERIAL=2 1 13 14 TO 4 16 1 1.

TRUSS MATERIAL=3 1 18 19 TO 2

19 1 1.

TRUSS MATERIAL=4 112 1 1.

17 20 B29.5

SOLVIA Verification Manual SOLVIA-PRE input cont.)

EGROUP 5

TRUSS MATERIAL=5 ENODES

/

1 2 3

TO 10 11 12 EDATA

/

1 1.

T-BOUNDARY SEGMENTS RAD-MATERIAL=6 INPUT=NODES 1 1 1. 1.

T-LOADS HEATFLOW 1 0.136E6 1 T-LOADS ENVIRONMENT RADIATION=YES INPUT=NODES 1 1.

2 SET VIEW=-Y PLOTORIENTATION=PORTRAIT NSYMBOL=YES, NNUMBERS=YES ENUMBERS=GROUP SUBFRAME 13 MESH ZONENAME=EG1 MESH ZONENAME=EG2 MESH ZONENAME=EG3 SUBFRAME 13 MESH ZONENAME=EG4 MESH ZONENAME=EG5 MESH NNUMBERS=NO ENUMBERS=NO SOLVIA-TEMP END SOLVIA-POST input B29 TRANSIENT TEMPERATURE ANALYSIS OF A SPACE SHUTTLE T-DATABASE CREATE WRITE FILENAME='b29.1is' SUBFRAME 21 NHISTORY NODE=i KIND=TEMPERATURE OUTPUT=ALL NHISTORY NODE=I KIND=TFLOW END Version 99.0 Nonlinear Examples B29.6

v t e Letter Sequence Response to RAI
TAC:MB3603, (Open) TAC:MB3604, (Open)
Initiation Request, Request, Request, Request, Request, Request, Request Acceptance... Supplement Administration Questions 31 May 2002 Answers 13 March 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 16 July 2002 Results Approval... Other: L-PI-03-016, Response to Opportunity for Comment on Task Interface Agreement (TIA) 2001-10, Design Basis Assumptions for Ability of Prairie Island, Unit 2, Emergency Diesel Generators to Meet Single-Failure Criteria for External Events, ML013480323, ML020020074, ML020020108, ML020030002, ML020410002, ML020570514, ML021060641, ML021070317, ML021140405, ML021140426, ML021610096, ML021990405, ML022140006, ML022600292, ML023290377, ML023640310, ML031010026, ML032040412, ML050900176, ML051030051
MONTHYEAR2002-05-28 [Table View]

ML022030415

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